Important Questions

Punjab State Board PSEB 12th Class Physics Important Questions Chapter 9 Ray Optics and Optical Instruments Important Questions and Answers.

PSEB 12th Class Physics Important Questions Chapter 9 Ray Optics and Optical Instruments

Very short answer type questions

Question 1.
What is reflection?
Answer:
When a light ray incident on a smooth surface bounces back to the same medium, it is called reflection.

Question 2.
State new cartesian sign conventions used for mirrors.
Answer:

• All the distances are measured from the pole of the mirror.
• All the distances measured in the direction of incident ray are taken as positive and the distances measured opposite to the incident ray are taken as – ve.
• All heights measured perpendicular to the principal axis in the upward direction are taken as + ve and those measured in downward direction are taken as – ve.

Note: Direction of incident light is always to be shown falling from left to right. So distance of the object and real image is always -ve while that of virtual image is always + ve, height of real image is always – ve while that of the virtual image and the size of real object are always + ve.

Question 3.
How does focal length of a lens change when red light incident on it is replaced by violet light? Give reason for your answer.
Answer:
The refractive index of the material of a lens increases with the decrease in wavelength of the incident light. So, focal length will decrease with a decrease in wavelength according to the formula.
\(\frac{1}{f}=(\mu-1)\left(\frac{1}{R_{1}}-\frac{1}{R_{2}}\right)\)
Thus, when we replace red light with violet light then due to increase in wavelength the focal length of the lens will decrease.

Question 4.
Define refraction of light.
Answer:
It is defined as the process of bending of light from its path when it travels from one medium to the another.

Question 5.
State
(a) Laws of reflection.
(b) Laws of refraction.
Answer:
(a) The following are the two laws of reflection :
(i) Angle of incidence is always equal to the angle of reflection.
(ii) The incident ray, reflected ray and normal to the surface at the point of incidence all lie in the same plane.

(b) The following are the two laws of refraction :
(i) The ratio of the sine of angle of incidence to the sine of the angle of refraction is always constant for a given pair of media.
i.e., \(\frac{\sin i}{\sin r}\) = constant = ^aµb
where ^aµb is called relative refractive index of medium b w.r.t. a.
(ii) The incident ray, refracted ray and the normal to the refracting surface at the point of incidence all lie in the same plane.

Question 6.
(i) What is the relation between critical angle and refractive index of a material?
(ii) Does critical angle depend on the colour of light? Explain.
Answer:
(i) Refractive index (µ) = \(\frac{1}{\sin C}\)
where, C is the critical angle.
(ii) Since, refractive index depends upon the wavelength of light, the critical angle for a given pair of media is different for different wavelengths (colours) of light.

Question 7.
Under what condition does a biconvex lens of glass having a certain refractive index act as a plane glass sheet when immersed in a liquid?
Answer:
A biconvex lens will act like a plane sheet of glass if it is immersed in a liquid having the same index of refraction as itself. In this case, the focal length 1/f = 0 or f→ ∞.

Question 8.
A biconvex lens made of a transparent material of refractive index 1.25 is immersed in water of refractive index 1.33. Will the lens behave as a converging lens? Give reason.
Answer:
No, it will behave as a diverging lens.
On Using thin lens maker formula
\(\frac{1}{f_{w}}=\left(\frac{n_{g}}{n_{m}}-1\right)\left(\frac{1}{R_{1}}-\frac{1}{R_{2}}\right)\)
On Using sign convention R₁ = +ve, R₂ = -ve and n_g = 1.25 and n_m = 1.33
\(\left(\frac{1}{R_{1}}+\frac{1}{R_{2}}\right)\) +ve,value and \(\left(\frac{1.25}{1.33}-1\right)\) =-ve value Hence f_w = -ve , so it behaves as a diverging lens.

Question 9.
Define total internal reflection.
Answer:
It is defined as the process of reflection of light that takes place when a ray of light travelling from denser to rarer medium gets incident at the interface of the two media at an angle greater than the critical angle for the given air of media.

Question 10.
State the criteria for the phenomenon of total internal reflection of light to take place.
Answer:
Following are the criteria for total internal reflection

• Light must pass from a denser to a rarer medium.
• Angle of incidence must be greater than critical angle.

Question 11.
Define mirage.
Answer:
It is defined as an optical illusion that occurs in deserts and coal tarred roads appear to be covered with water but on approaching at that place no water is obtained. In deserts thirsty animals observe virtual images of trees on hot sand so expecting a pond of water there but on reaching there, they do not get water pond and hence called optical illusion.

Question 12.
Why diamond sparkles?
Answer:
The critical angle for diamond is low i.e., 23° and its refractive index is 2.47. The faces of diamond are cut in such a way that when a ray of light entering from a face undergoes multiple total internal reflections from its different faces. Due to small value of the critical angle, almost all light rays entering the diamond suffer multiple total internal reflection and thus it shines brilliantly.

Question 13.
What are optical fibres? Give their one use.
Answer:
Optical fibres are thousands of very fine quality fibres of glass or quartz. The diameter of each fibre is of the order of 10^-4 cm having refractive index of material equal to 1.7. These are coated with a thin layer of material having µ = 1.5.
They are used in transmission and reception of electrical signals by converting them first into light signals.

Question 14.
Write the relationship between angle of incidence ‘i’ angle of prism ‘A’ and angle of minimum deviation for a triangular prism.
Answer:
i = \(\frac{A+\delta_{m}}{2}\)
where, δ_m = angle of minimum deviation.

Question 15.
Define dispersion of light. What is its cause?
Answer:
It is defined as the process of splitting up of white light into its constituent colours on passing through a prism.
We know that for small angled prism,
δ = (µ -1)A.
Also according to Cauchy’s formula, we know that µ ∝ \(\frac{1}{\lambda^{2}}\)
Thus µ of the material of prism is different for different colours, so δ is also different for different incident colours.
Thus due to different values of angle of deviation, each colour occupies different direction in emergent beam of light and thus constituent colours of white light get dispersed. λv < λr, so δv > δr.
The violet colour deviates more than the red colour.

Question 16.
Explain the rainbow.
Answer:
The rainbow is an example of the dispersion of sunlight by the water drops in the atmosphere. This is a phenomenon due to combined effect of dispersion, refraction and reflection of sunlight by spherical water droplets of rain. The conditions for observing a rainbow are that the sun should be shining in one part of the sky (say near western horizon) while it is raining in the opposite part of the sky (say eastern horizon). An observer can therefore see a rainbow only when his back is towards the sun.

Question 17.
Why does the Sun look reddish at sunset or sunrise?
Answer:
During sunset or sunrise, the sun is just above the horizon, the blue colour gets scattered most by the atmospheric molecules while red light gets scattered least, hence Sun appears red.
I ∝ \(\frac{1}{2^{4}}\) and λ_B << λ_R.

Question 18.
Will the focal length of a lens for red light be more, same or less than that for blue light? (NCERT Exemplar)
Answer:
As the refractive index for red is less than that for blue, parallel beams of light incident on a lens will be bent more towards the axis for blue light compared to red. Thus the focal length for red light will be more than that for blue.

Question 19.
An unsymmetrical double convex thin lens forms the image of a point object on its axis. Will the position of the image change if the lens is reversed? (NCERT Exemplar)
Answer:
No, the reversibility of the tens makes equation.
\(\frac{1}{v}-\frac{1}{u}=\frac{1}{f}=(n-1)\left(\frac{1}{R_{1}}-\frac{1}{R_{2}}\right)\)
= -(n-1) \(\left(\frac{1}{R_{2}}-\frac{1}{R_{1}}\right)\)
On reversing the lens, values of R₁ and R₂ are reversed and so their signs.
Hence, for a given position of object (u), position of image (v) remains unaffected.

Question 20.
Why danger signals are of red light?
Answer:
Scattering of light is inversely proportional to the fourth power of wavelength of incident light. As red light has longer wavelength as compared to other visible colours, so its scattering is least and thus red light signals can be seen from a longer distance.

Short answer type questions

Question 1.
Will the focal length of a lens for red light be more, same or less than that for blue light? [NCERT Exemplar]
Answer:
As the refractive index for red is less than that for blue parallel beams of light incident on a lens will be bent more towards the axis for blue light compared to red.
In other words, μ_b > μ_r By lens maker’s formula,
\(\frac{1}{f}=\left(n_{21}-1\right)\left(\frac{1}{R_{1}}-\frac{1}{R_{2}}\right)\)
Therefore, f_b < f_r
Thus, the focal length for blue light will be smaller than that for red.

Question 2.
Define power of a lens. Write its units. Deduce the relation \(\frac{1}{f}=\frac{1}{f_{1}}+\frac{1}{f_{2}}\) for two thin lenses kept in contact coaxially.
Answer:
The power of a lens is equal to the reciprocal of its focal length when it is measured in metre. Power of a lens,
P = \(\frac{1}{f(\text { metre })}\)
Its SI unit is dioptre (D).

Consider two lenses A and B of focal lengths, f₁ and f₂ placed in contact with each other. An object is placed at a point O beyond the focus of the first lens A. .
The first lens produces an image (real image) at I₁ which serves as a virtual object for the second lens B producing the final image at I.

Since, the lenses are thin, we assume the optical centres P of the lenses to be coincident. For the image formed by the first lens A, we obtain
\(\frac{1}{v_{1}}-\frac{1}{u}=\frac{1}{f_{1}}\) ……………………………. (1)
For the image formed by the second lens B, we obtain
\(\frac{1}{v}-\frac{1}{v_{1}}=\frac{1}{f_{2}}\) …………………………………. (2)
Adding eqs. (1) and (2), we obtain
\(\frac{1}{v}-\frac{1}{u}=\frac{1}{f_{1}}+\frac{1}{f_{2}}\) …………………………. (3)

If the two lenses system is regarded as equivalent to a single lens of focal length f, we have
\(\frac{1}{v}-\frac{1}{u}=\frac{1}{f}\) ……………………………… (4)
From eqs. (3) and (4), we obtain
\(\frac{1}{f_{1}}+\frac{1}{f_{2}}=\frac{1}{f}\) .

Question 3.
(a) Draw a schematic labelled ray diagram of a reflecting type telescope (cassegrain).
(b) The objective of telescope is of larger focal length and of larger aperture (compared to the eyepiece). Why? Given reasons.
(c) State the advantages of reflecting telescope over refracting telescope.
Answer:
(a)

(b) In normal adjustment, magnifying power of the telescope, M = \(\frac{f_{0}}{f_{e}}\)
(i) If focal length of the objective lens is large in comparison to the eyepiece, magnifying power increases.
(ii) Resolving power of the telescope RP = \(\frac{D}{1.22 \lambda}\)
D being the diameter of the objective. To increase the resolving power of the telescope, large aperture of the objective lens is required.

Advantages

• There is no chromatic aberration in a mirror.
• Brighter image.
• High resolving power.
• Large magnifying power.

Question 4.
How is the working of a telescope different from that of a microscope?
Answer:
Difference in working of telescope and microscope

• Objective of telescope forms the image of a very far off object at or within the focus of its eyepiece. The microscope does the same for a small object kept just beyond the focus of its objective.
• The final image formed by a telescope is magnified relative to its size as seen by the unaided eye while the final image formed by a microscope is magnified relative to its absolute size.
• The objective of a telescope has large focal length and large aperture while the corresponding parameters for a microscope have very small ‘ values.

Question 5.
For a glass prism (μ = \(\sqrt{3}\) ) the angle of minimum deviation is equal to the angle of the prism. Find the angle of the prism. (NCERT Exemplar)
Answer:
At minimum deviation μ = \(\frac{\sin \left[\frac{\left(A+\delta_{m}\right)}{2}\right]}{\sin \left(\frac{A}{2}\right)} \)
Given, δ_m = A
∴ μ = \(\frac{\sin A}{\sin \frac{A}{2}}=\frac{2 \sin \frac{A}{2} \cos \frac{A}{2}}{\sin \frac{A}{2}}=2 \cos \frac{A}{2}\)
∴ \(\cos \frac{A}{2}=\frac{\sqrt{3}}{2} \text { or } \frac{A}{2}=30\)
⇒ A = 60⁰.

Long Answer Type Questions

Question 1.
(a) Draw a ray diagram for formation of image of a point object by a thin double convex lens having radii of curvature R₁ and R₂. Hence, derive lens maker’s formula for a double convex lens. State the assumptions made and sign convention used.
(b) A convex lens is placed over a plane mirror. A pin is now positioned so that there is no parallax between the pin and its image formed by this lens-mirror combination. How will you use this observation to find focal length of the lens? Explain briefly.
Answer:
(a) Lens Maker’s Formula: Suppose L is a thin lens. The refractive index of the material of lens is n₂ and it is placed in a medium of refractive index n₁. The optical centre of lens is C and X’ X is principal axis. The radii of curvature of the surfaces of the lens are R₁ and R₂ and their poles are P₁ and P₂.

The thickness of lens is t, which is very small. O is a point object on the principal axis of the lens. The distance of O from pole P₁ is u. The first refracting surface forms the image of O at I’ at a distance v’ from P₁.
From the refraction formula at spherical surface, \(\frac{n_{2}}{v^{\prime}}-\frac{n_{1}}{u}=\frac{n_{2}-n_{1}}{R_{1}}\) ……………………………….. (1)
The image I’ acts as a virtual object for second surface and after refraction at second surface, the final image is formed at I.

The distance of I from pole P₂ of second surface is v. The distance of virtual object (I’) from pole P₂ is (v’ -t).
For refraction at second surface, the ray is going from second medium (refractive index n₂) to first medium (refractive index n₁), therefore, from refraction formula at spherical surface

\(\frac{n_{1}}{v}-\frac{n_{2}}{\left(v^{\prime}-t\right)}=\frac{n_{1}-n_{2}}{R_{2}}\) ……………….. (2)
For a thin lens t is negligible as compared to v’, therefore from eq. (2)
\(\frac{n_{1}}{v}-\frac{n_{2}}{v^{\prime}}=-\frac{n_{2}-n_{1}}{R_{2}}\) ……………………………….. (3)
Adding equation (1) and (3),we get

where, ₁n₂ = \(\frac{n_{2}}{n_{1}}\) is refractive index of second medium (te. medium of lens) with respect to first medium. If the object O is at infinity, the image will be formed at second focus i.e., if u = ∞, v = f₂ =f
Therefore, from equation (4)

This formula is called Lens-Maker’s formula. If first medium is air and refractive index of material of lens be n, then ₁n₂ = n, therefore, the modified equation (5) may be written as
\(\frac{1}{f}=(n-1)\left(\frac{1}{R_{1}}-\frac{1}{R_{2}}\right)\) ………………………………. (6)

(b) Focal length = distance of the pin from the mirror.
The rays from the object after refraction from lens should fall normally on the plane mirror. So, they retrace their path. Hence, rays must be originating from focus and thus distance of the pin from the plane mirror gives focal length of the lens.

Question 2.
(a) Draw the labelled ray diagram for the formation of image by a compound microscope. Derive an expression for its total magnification (or magnifying power), when the final image is formed at the near point. Why both objective and eyepiece of a compound microscope must have short focal lengths?
(b) Draw a ray diagram showing the image formation by a compound microscope. Hence, obtain expression for total magnification when the image is formed at infinity.
Answer:
(a) Compound Microscope: It consists of a long cylindrical tube, containing at one end a convex lens of small aperture and small focal length. This is called the objective lens (0). At the other end of the tube another co-axial smaller and wide tube is fitted, which carries a convex lens (E) at its outer end. This lens is towards the eye and is called the eyepiece. The focal length and aperture of eyepiece are somewhat larger than those of objective lens. Cross-wires are mounted at a definite distance before the eyepiece. The entire tube can be moved forward and backward by the rack and pinion arrangement.

Adjustment: First of all the eyepiece is displaced backward and forward to focus it on cross-wires. Now the object is placed just in front of the objective lens and the entire tube is moved by rack and pinion arrangement until there is no parallax between image of object and cross wire. In this position, the image of the object appears quite distinct.

Working: Suppose a small object AB is placed slightly away from the first focus F_o‘of the objective lens. The objective lens forms the real, inverted and magnified image A’ B’, which acts as an object for eyepiece. The eyepiece is so adjusted that the image A’B’ lies between the first focus F_e‘ and the eyepiece E. The eyepiece forms its image A”B” which is virtual, erect and magnified. Thus the final image A”B” formed by the microscope is inverted and magnified and its position is outside the objective and eyepiece towards objective lens.

The magnifying power of a microscope is defined as the ratio of angle (β) subtended by final image on the eye to the angle (α) subtended by the object on eye, when the object is placed at the least distance of distinct vision, i.e., Magnifying power,
M = \(\frac{\beta}{\alpha}\)
As object is very small, angles a and 1 are very small and so tan α = α and tan β = β. By definition the object AB is placed at the least distance of distinct vision.

∴ α = tan α = \(\frac{A B}{E A}\)
By sign convention, EA = – D,
∴ α = \(\frac{A B}{-D}\)
and from figure
β = tan β = \(\frac{A^{\prime} B^{\prime}}{E A^{\prime}}\)
If u_e is distance of image A’ B’ from eyepiece E, then by sign convention, EA’ = -u_e
and so, β = \(\frac{A^{\prime} B^{\prime}}{-u_{e}}\)

Hence, magnifying power,
M = \(\frac{\beta}{\alpha}=\frac{A^{\prime} B^{\prime} /\left(-u_{e}\right)}{A B /(-D)}=\frac{A^{\prime} B^{\prime}}{A B} \cdot \frac{D}{u_{e}}\)
By sign conventions, magnification of objective lens
\(\frac{A^{\prime} B^{\prime}}{A B}=\frac{v_{o}}{\left(-u_{o}\right)}\)
∴ M = \(-\frac{v_{o}}{u_{o}} \cdot \frac{D}{u_{e}}\) ………………………………….. (2)

Using lens formula \(\frac{1}{f}=\frac{1}{v}-\frac{1}{u}\) for eyelens,
(i.e. using f = f_e’ V = -v_e, U = -u_e ) we get
\(\frac{1}{f_{e}}=\frac{1}{-v_{e}}-\frac{1}{\left(-u_{e}\right)}\)
or \(\frac{1}{u_{e}}=\frac{1}{f_{e}}+\frac{1}{v_{e}}\)
Magnifying power,
M = \(-\frac{v_{o}}{u_{o}} D\left(\frac{1}{f_{e}}+\frac{1}{v_{e}}\right)\)

or M = \(-\frac{v_{o}}{u_{o}}\left(\frac{D}{f_{e}}+\frac{D}{v_{e}}\right)\)
When final image is formed at the distance of distinct vision, V_e = D
∴ Magnification,
M= – \(\frac{v_{o}}{u_{o}}\left(1+\frac{D}{f_{e}}\right)\)

For greater magnification of a compound microscope, f_e should be small. As f_o < f_e’ so f₀ is small.
Hence, for greater magnification both f₀ and f_e should be small with f₀ to be smaller of the two.

(b) If image A’B’ is exactly at the focus of the eyepiece, then image A”B” is formed at infinity.
If the object AB is very close to the focus of the objective lens of focal length f₀, then magnification M₀ by the objective lens
M_e = \(\frac{L}{f_{0}}\)

where, L is tube length (or distance between lenses L₀ and L_e) Magnification M_e by the eyepiece
M_e = \(\frac{D}{F_{e}} \)
where, D = Least distance of distinct vision
Total magnification, m = M₀ M_e = \(\left(\frac{L}{f_{o}}\right)\left(\frac{D}{f_{e}}\right)\)

Question 3.
Explain with the help of a labelled ray diagram, how is image formed in an astronomical telescope. Derive an expression for its magnifying power.
Or
Draw a ray diagram showing the image formation of a distant object by a refracting telescope. Define Its magnifying power and write the two important factors considered to increase the magnifying power. Describe briefly the two main limitations and explain how far these can be minimised in a reflecting telescope.
Answer:
Astronomical (Refracti ng) Telescope
Construction: It consists of two co-axial cylindrical tubes, out of which one tube is long and wide, while the other tube is small and narrow. The narrow tube may be moved in and out of the wide tube by rack and pinion arrangement. At one end of wide tube an achromatic convex lens L₁ is placed, which faces the object and is so-called objective (lens). The focal length and aperture of this lens are kept large. The large aperture of objective is taken that it may collect sufficient light to form a bright image of a distant object. The narrow tube is towards eye and carries an achromatic convex lens 12 of small focal length and small aperture on its outer end. This is called eye-lens or eyepiece.

The small aperture of eye lens is taken so that the whole light refracted by it may reach the eye. Cross-wires are fitted at a definite distance from the eye lens. Due to large focal length of objective lens and small focal length of eye lens, the final image subtends a large angle at the eye and hence the object appears large. The distance between the two lenses may be arranged by displacing narrow tube in or out of wide tube by means of rack and pinion arrangement.

Adjustment: First of all the eyepiece is moved backward and forward in the narrow tube and focused on the cross-wires. Then the objective lens is directed towards the object and narrow tube is displaced in or out of wide tube until the image of object is formed on cross-wires and there is no parallax between the image and cross-wires. In this position, a clear image of the object is seen. As the image is formed by refraction of light through both the lenses, this telescope is called the refracting telescope.

Working: Suppose AB is an object whose end A is on the axis of telescope. The objective lens (L₁) forms the image A’B’ of the object AB at its second principal focus F₀.
This image is real, inverted and diminished. This image A’ B’ acts as an object for the eyepiece L₂ and lies between first focus f_e‘ and optical centre C₂ of lens L₂.
Therefore, eyepiece forms its image A” B” which is virtual, erect and magnified.
Thus, the final image A” B” of object AB formed by the telescope is magnified, inverted and lies between objective and eyepiece.

Magnifying Power: The magnifying power of a telescope is measured by the ratio of angle (β) subtended by final image on the eye to the angle (α) subtended by object on the eye. i.e.,
Magnifying power M = \(\frac{\beta}{\alpha}\)
As α and β are very small angles, therefore, from figure.

The angle subtended by final image A” B” on eye.
β = angle subtended by image A’B’ on eye
= tanβ = \(\frac{A^{\prime} B^{\prime}}{C_{2} A^{\prime}}\)
As the object is very far (at infinity) from the telescope, the angle subtended by object at eye is same as the angle subtended by object on objective lens.
∴ α = tan α = \(\frac{A^{\prime} B^{\prime}}{C_{1} A^{\prime}}\)
∴ M = \(\frac{\beta}{\alpha}=\frac{A^{\prime} B^{\prime} / C_{2} A^{\prime}}{A^{\prime} B^{\prime} / C_{1} A^{\prime}}=\frac{C_{1} A^{\prime}}{C_{2} A^{\prime}}\)
If the focal lengths of objective and eyepiece be f₀ and f_e, distance of image A’B’ from eyepiece be u_e, then by sign convention
C₁A’ = +f₀
C₂A’ = – u_e
∴ M = –\(\frac{f_{o}}{u_{e}}\) ……………………………… (1)

If v_e is the distance of A” B” from eye-piece, then by sign convention, f_e is positive, u_e and v_e are both negative. Hence, by lens formula = \(\frac{1}{f}=\frac{1}{v}-\frac{1}{u}\)
we have
\(\frac{1}{f_{e}}=\frac{1}{-v_{e}}-\frac{1}{\left(-u_{e}\right)}\)
or
\(\frac{1}{u_{e}}=\frac{1}{f_{e}}+\frac{1}{v_{e}}\)
Substituting this value in eq. (1), we get
M = -f₀ \(\left(\frac{1}{f_{e}}+\frac{1}{v_{e}}\right)\) …………………………. (2)

This is the general formula for magnifying power. In this formula, only numerical values of f₀, f_e and v_e are to be used because signs have already been used.
Length of Telescope : The distance between objective and eyepiece is called the length (L) of the telescope. Obviously,
L = L₁L₂ =C₁C₂ = f₀+u_e …………………… (3)

Now there arise two cases :
(i) When the final image is formed at minimum distance (D) of distinct vision then v_e =D
∴ M = -f₀ \(\left(\frac{1}{f_{e}}+\frac{1}{D}\right)=-\frac{f_{o}}{f_{e}}\left(1+\frac{f_{e}}{D}\right)\) …………………………… (4)
Length of telescope L = f₀ + u_e

(ii) In normal adjustment position, the final image is formed at infinity: For relaxed eye, the final image is formed at infinity. In this state, the image A’B’ formed by objective lens should be at first the principal focus of eyepiece, i.e.,
u_e = f_e and v_e
∴ Magnifying power,
M = – f₀ \(\left(\frac{1}{f_{e}}+\frac{1}{\infty}\right)\) = –\(\frac{f_{o}}{f_{e}}\)
Length of telescope = f₀ + f_e
For large magnifying power, f₀ should be large and f_e should be small. For high resolution of the telescope, diameter of the objective should be large.

Factors for Increasing the Magnifying Power
1. Increasing focal length of objective
2. Decreasing focal length of eyepiece

Limitations
1. Suffers from chromatic aberration
2. Suffers from spherical aberration
3. Small magnifying power
4. Small resolving power

Advantages of Reflecting Telescope
1. No chromatic aberration, because mirror is used.
2. Spherical aberration can be removed by using a parabolic mirror.
3. Image is bright because no loss of energy due to reflection.
4. Large mirror can provide easier mechanical support.

Punjab State Board PSEB 12th Class Physics Important Questions Chapter 6 Electromagnetic Induction Important Questions and Answers.

PSEB 12th Class Physics Important Questions Chapter 6 Electromagnetic Induction

Very short answer type questions

Question 1.
The motion of copper plate is damped when it is allowed to oscillate between the two poles of a magnet. What is the cause of this damping?
Answer:
As the plate oscillate, the changing magnetic flux through the plate produces a strong eddy current in the direction, which opposes the cause. Also, copper being substance, it gets magnetised in the opposite direction, so the plate motion gets damped.

Question 2.
On what factors does the magnitude of the emf induced in the circuit due to magnetic flux depend ?
Answer:
Depends on the time rate of change in magnetic flux (or simply change in Magnetic flux)
\(|\varepsilon|=\frac{\Delta \phi}{\Delta t}\)

Question 3.
A light metal disc on the top of an electromagnet is thrown up as the current is switched on. Why? Give reason.
Answer:
When the current begins to grow through the electromagnet, the magnetic flux through the disc begins to increase. This sets up eddy current in the disc in the same direction as that of the electromagnetic current.

Thus, if the upper surface of electromagnetic acquires AT-polarity, the lower surface of the disc also acquires N-polarity. As, same magnetic poles repel each other, the light metallic disc is thrown up.

Question 4.
State the Faraday’s law’ of electromagnetic induction.
Answer:
On the basis of his experiment, Faraday gave the following two laws:
First Law: Whenever magnetic flux linked with a circuit changes, an emf is induced in it which lasts, so long as change in flux continuous.
Second Law: The emf induced in loop or closed circuit is directly proportional to the rate of change of magnetic flux linked with the loop
i.e., ε ∝ \(\frac{(-) d \phi}{d t}\) or ε = -N \(\frac{d \phi}{d t}\)
where, N= number of turns in the coil. Negative sign indicates the Lenz’s law.

Question 5.
State Lenz’s law. A metallic rod held horizontally along East-West direction, is allowed to fall under gravity. Will there be an emf induced at its ends? Justify your answer.
Answer:
Lenz’s Law: The direction of the induced emf, or the current, in any circuit is such as to oppose the cause that produces it.

Yes, emf will be induced in the rod as there is change in magnetic flux. When a metallic rod held horizontally along East-West direction, is allowed to fall freely under gravity i.e., fall from North to South, the intensity of magnetic lines of the earth’s magnetic field changes through it, i.e., the magnetic flux changes and hence emf induced in it.

Question 6.
How does the mutual inductance of a pair of coils change, when
(i) distance between the coils is increased and
(ii) number of turns in the coils is increased?
Answer:
(i) AΦ = MI, with the increase in the distance between the coils the magnetic flux linked with the secondary coil decreases and hence, the mutual inductance of the two coils will decreases with the increase of separation between them.

(ii) Mutual inductance of two coils can be found out by
M = μ₀N₁N₂ Al i.e.,
M ∝ N₁N₂, SO, with the increase in number of turns mutual inductance increases.

Question 7.
Why is the core of a transformer laminated?
Answer:
The core of a transformer is laminated because of preventing eddy current being produced in the core.

Question 8.
How can the self-inductance of a given coil having N number of turns, area of cross-section A and lengths l be increased?
Answer:
The self-inductance can be increased by the help of electric fields. It does not depend on the current through circuit but depends upon the permeability of material from which the core is made up off.

Question 9.
Consider a magnet surrounded by a wire with an on/off switch S (as shown in figure). If the switch is thrown from the off position (open circuit) to the on position (closed circuit), will a current ?Explain (NCERT Exemplar)

Answer:
No part of the wire is moving and so motional e.m.f. is zero. The magnet is stationary and hence the magnetic field does not change with time. This means no electromotive force is produced and hence no current will flow in the circuit.

Question 10.
A wire in the form of a tightly wound solenoid is connected to a DC source, and carries a current. If the coil is stretched so that there are gaps between successive elements of the spiral coil, will the current increase or decrease? Explain. (NCERT Exemplar)
Answer:
The current will increase. As the wires are pulled apart the flux will leak through the gaps. Lenz’s law demands that induced emf resist this decrease, which can be done by an increase in current.

Question 11.
A solenoid is connected to a battery so that a steady current flows through it. If an iron core is inserted into the solenoid, will the current increase or decrease? Explain. (NCERT Exemplar)
Answer:
The current will decrease. As the iron core is inserted in the solenoid, the magnetic field increases and the flux increases. Lenz’s law implies that induced emf should resist this increase, which can be achieved by a decrease in current. However, this change will be momentarily.

Question 12.
Consider a metal ring kept (supported by a cardboard) on top of a fixed solenoid carrying a current /. The centre of the ring coincides with the axis of the solenoid. If the current in the solenoid is switched off, what will happen to the ring? (NCERT Exemplar)
Answer:
When the current in the solenoid decreases a current flows in the same direction in the metal ring as in the solenoid. Thus there will be a downward force. This means the ring will remain on the cardboard. The upward reaction of the cardboard on the ring will increase.

Short answer type questions

Question 1.
Consider a closed loop C in a magnetic field (figure). The flux passing through the loop is defined by choosing a surface whose edge coincides with the loop and using the formula Φ = B₁ dA₁, B₂ dA₂…. Now, if we choose two different surfaces S₁ and S₂ having C as their edge, would we get the same answer for flux. Justify your answer. (NCERTExemplar)

The magnetic flux linked with the surface can considered as the number of magnetic field lines passing through the surface. So, let dΦ = BdA represents magnetic lines in an area A to B.

By the concept of continuity of lines B cannot end or start in space, therefore the number of lines passing through surface S₁ must be the same as the number of lines passing through the surface S₂. Therefore, in both the cases we gets the same answer for flux.

Question 2.
What are eddy currents? Write their two applications.
Answer:
Eddy Current: Eddy currents are the currents induced in the bulk pieces of conductors when the amount of magnetic flux linked with the conductor changes.

Eddy currents can be minimised by taking laminated core, consists of thin metallic sheet insulated from each other by varnish instead of a single solid mass. The plane of the sheets should be kept perpendicular to the direction of the currents. The insulation provides high resistance hence, eddy current gets minimised.

Applications
(i) Electromagnetic damping
(ii) Induction furnace.

Question 3.
(i) A rod of length l is moved horizontally with a uniform – velocity v in a direction perpendicular to its length through a region in which a uniform magnetic field is acting vertically downward. Derive the expression for the emf induced across the ends of the rod.

(ii) How does one understand this motional emf by invoking the Lorentz force acting on the free charge carriers of the conductor? Explain.
Answer:
(i) Consider a straight conductor moving with velocity v and U shaped conductor placed in perpendicular magnetic field as shown in the figure.

Let conductor shifts from ab to a’ b’ in time dt, then change in magnetic flux
dΦ = B × change in area
= B × (areaa’b’ab)
= B × (l × vdt)
∴  \(\frac{d \phi}{d t}\) Bvl
∴  Induced emf lei \(|\varepsilon|=\frac{d \phi}{d t}\) = Bvl

(ii) During motion, free e^– are shifted at one end due to magnetic force so due to polarisation of rod electric field is produced which applies electric force on free e^– on opposite direction.

At equilibrium of Lorentz force,
F_e + F_m = 0
qE + q(v × B) = 0
E = -v × B = B × v
\(|E|=|B v \sin 90|\)
\(\frac{d v}{d r}\) = Bv
P_D = Bvl

Question 4.
(a) How does the mutual inductance of a pair of coils change when
(i) distance between the coils is increased and
(ii) number of turns in the coils is increased?

(b) A plot of magnetic flux (Φ) versus current (I), is shown in the figure for two inductors A and B. Which of the two has large value of self-inductance?

(c) How is the mutual inductance of a pair of coils affected when
(i) separation between the coils is increased?
(ii) the number of turns in each coil is increased?
(iii) a thin iron sheet is placed between the two coils, other factors remaining the same?
Justify your answer in each case.

Answer:
(a)
(i) Mutual inductance decreases.
(ii) Mutual inductance increases.
Concept
(i) If distance between two coils is increased as shown in figure.

It causes decrease in magnetic flux linked with the coil C₂. Hence induced emf in coil C₂ decreases by relation ε₂ = \(\frac{-d \phi_{2}}{d t}\). Hence mutual inductance decreases.
(ii) From relation M₂₁ = μ₀N₁N₂ Al, if number of turns in one of the coils or both increases, means mutual inductance will increase.

(b) Φ = LI ⇒ \(\frac{\phi}{I}\) = L
The slope of \(\frac{\phi}{I}\) of straight line is equal to self-inductance L. It is larger for inductor A; therefore inductor A has larger value of self inductanc ‘ L’.

(c)
(i) When the relative distance between the coil is increased, the leakage
of flux increases which reduces the magnetic coupling of the coils. So magnetic flux linked with all the turns decreases. Therefore, mutual inductance will be decreased.

(ii) Mutual inductance for a pair of coil is given by
M = K\(\sqrt{L_{1} L_{2}}\)
where, L = \(\frac{\mu N^{2} A}{l}\) and L is called self inductance. Therefore, when the number of turns in each coil increases, the mutual inductance also increases.

(iii) When a thin iron sheet is placed between the two coils, the mutual inductance increases because M ∝ permeability. The permeability of the medium between coils increases.

Question 5.
Consider a metallic pipe with an inner radius of 1 cm. If a cylindrical bar magnet of radius 0.8 cm is dropped through the pipe, It takes more time to come down than It takes for a similar
unmagnetised cylindrical iron bar dropped through the metallic pipe. Explain. (NCERT Exemplar)
Answer:
For the magnet, eddy currents are produced in the metallic pipe. These currents will oppose the motion of the magnet. Therefore magnet’s downward acceleration will be less than the acceleration due to gravity g. On the other hand, an unmagnetised iron bar will not produce eddy
currents and will fall an acceleration g. Thus the magnet will take more time.

Question 6.
A magnetic field B = B₀ sin(ωt) k̂ covers a large region where a wire AB slides smoothly over two parallel conductors separated by a distance d (figure). The wires are in the x-y plane. The wire AB (of length d) has resistance R and the parallel wires have negligible resistance. If AB is moving with velocity y, what is the current in the circuit. What is the force needed to keep the wire moving at constant velocity? (NCERT Exemplar)

Answer:
Let us assume that the parallel wires are at y = 0 i. e., along x-axis and y = d. At t = 0, AB has x = 0, i. e., along y-axis and moves with a velocity v. Let at time t, wire is at x (t) = vt.
Now, the motional emf across AB is
= (B₀sinωt) vd(-ĵ)
emf due to change in field (along OBAC)
= -B₀ωcosωt (t)d
Total emf in the circuit = emf due to change in field (along OBAC) + the motional emf across AB = -B₀d[ωxcos(ωt) + vsin (ωt)]
Electric current in clockwise direction is given by,
= \(\frac{B_{0} d}{R}\) = (ωxcosωt + vsinωt)
The force acting on the conductor is given by F = ilB sin 90° = ilB
Substituting the values, we have
Force needed along i = \(\frac{B_{0} d}{R}\) (ωx cos ωt + vsinωt) × d × B sinωt
= \(\frac{B_{0}^{2} d^{2}}{R}\)(ωx cos ωt + vsinωt) sinωt
This is the required expression for force.

Long answer type questions

Question 1.
(i) How is magnetic flux linked with the armature coil changed in a generator ?
(ii) Derive the expression for maximum value of the induced emf and state the rule that gives the direction of the induced emf.
(iii) Show the variation of the emf generated versus time as the armature is rotated with respect to the direction of the magnetic fields.
Answer:
(i) The direction of flow of current in resistance R get changed alternatively after every half cycle.
Thus, AC is produced in coil.

(ii) Let at any instant total magnetic flux linked with the armature coil is G. and θ = ωt is the angle made by area vector of coil with magnetic field.
Φ = NBA cosθ = NBA cosωt
\(\frac{d \phi}{d t}\) = -NBAω sin ωt
– \(\frac{d \phi}{d t}\) = NBAω sin ωt
By Faraday’s law of emf, e = \(\frac{-d \phi}{d t}\)
Induced emf in coil is given by,
e = NBAω sinωt
e = e₀ sinωt
where, e₀ = NBAω = peak value of induced emf

(iii) The mechanical energy spent in rotating the coil in magnetic field appears in the form of electrical energy.

Question 2.
State the working of AC generator with the help of a labelled diagram.
The coil of an AC generator having N turns, each of area A, is rotated with a constant angular velocity to. Deduce the expression for the alternating emf generated in the coil.
What is the source of energy generation in this device?
Answer:
AC Generator: A dynamo or generator is a device which converts mechanical energy into electrical energy.

Principle: It works on the principle of electromagnetic induction. When a coil rotates continuously in a magnetic field, the effective area of the coil linked normally with the magnetic field lines, changes continuously with time. This variation of magnetic flux with time results in the production of an alternating emf in the coil.

Construction: It consists of the four main parts
(i) Field magnet: It produces the magnetic field. In the case of a low power dynamo, the magnetic field is generated by a permanent magnet, while in the case of large power dynamo, the magnetic field is produced by an electromagnet.

(ii) Armature: It consists of a large number of turns of insulated wire in the soft iron drum or ring. It can revolve an axle between the two poles of the field magnet. The drum or ring serves the two purposes: (a) It serves as a support to coils and (b) It increases the magnetic field due to air core being replaced by an iron core.

(iii) Slip rings: The slip rings R₁ and R₂ are the two metal rings to which the ends of armature coil are connected. These rings are fixed to the shaft which rotates the armature coil so that the rings also rotate along with the armature.

(iv) Brushes: There are two flexible metal plates or carbon rods (B₁ and B₂) which are fixed and constantly touch the revolving rings. The output current in external load RL is taken through these brushes.

Working: When the armature coil is rotated in the strong magnetic field, the magnetic flux linked with the coil changes and the current is induced in the coil, its direction being given by Fleming’s right hand rule. Considering the armature to be in vertical position and as it rotates in anticlockwise direction, the wire ab moves upward and cd downward, so that the direction of induced current is shown in fig. In the external circuit, the current flows along B₁R_lB₂. The direction of current remains unchanged during the first half turn of armature. During the second half revolution, the wire ab moves downward and cd upward, so the direction of current is reversed and in external circuit it flows along B₂R_LB₁ Thus the direction of induced emf and current changes in the external circuit after each half revolution.

Expression for Induced emf: If N is number of turns in coil, f the frequency of rotation, A area of coil and B the magnetic induction, then induced emf
e = – \(\frac{d \phi}{d t}\) = –\(\frac{d}{d t}\) {NBA (cos 2π ft)} dt dt
= 2π NBA f sin 2π ft
Obviously, the emf produced is alternating and hence the current is also alternating.
Current produced by an AC generator cannot be measured by moving coil ammeter; because the average value of AC over full cycle is zero.
The source of energy generation is the mechanical energy of rotation of armature coil.

Punjab State Board PSEB 12th Class Physics Important Questions Chapter 10 Wave Optics Important Questions and Answers.

PSEB 12th Class Physics Important Questions Chapter 10 Wave Optics

Very short answer type questions

Question 1.
Define the term ‘wavefront’.
Answer:
It is defined as the locus of all points in a medium vibrating in the same phase.

Question 2.
State Huygen’s principle of diffraction of light.
Answer:
When a wavefront strikes to the corner of an obstacle, lightwave bends around the corner because every point on the wavefront again behaves like a . light source and emit secondary wavelets in all directions (Huygen’s wave theory) including the region of geometrical shadow. This explains diffraction.

Question 3.
Define the term ‘coherent sources’ which are required to produce interference patterns in Young’s double-slit experiment.
Answer:
Two monochromatic sources, which produce light waves, having a constant phase difference are known as coherent sources.

Question 4.
Define Doppler’s effect in light.
Answer:
It states that whenever there is a relative motion between the observer and the source of light, the apparent frequency of light received by the observer is different from the actual frequency of the light emitted by the source of light.

Question 5.
Define Doppler shift.
Answer:
It is defined as the apparent change in the frequency or wavelength of light due to the relative motion between the source and the observer.

Question 6.
Define redshift.
Answer:
It is defined as the shifting of radiations from the source of light towards the red end of the spectrum when the source moves away from the stationary observer. The wavelength increases due to redshift.

Question 7.
Define limit of resolution of an optical instrument.
Answer:
It is defined as the minimum distance by which the timepoint objects are separated so that their images can be seen as just separated by the optical instrument.

Question 8.
Define resolving power of the optical instruments.
Answer:
It is defined as the reciprocal of the limit of resolution of the optical instrument.

Question 9.
How are resolving power of a telescope change by increasing or decreasing the aperture of the objective?
Answer:
We know that the resolving power of telescope is given by
R.P. = \(\frac{D}{1.22 \lambda}\)
As R.p. ∝ D, so by increasing or decreasing D (aperture) of the objective, the resolving power is increased or decreased.

Question 10.
Which of the following waves can be polarised (i) Heat waves (ii) Sound waves? Give reason to support your answer.
Answer:
Heatwaves are transverse or electromagnetic in nature whereas sound waves are not. Polarisation is possible only for transverse waves.

Question 11.
How is linearly polarised light obtain by the process of scattering of light? find the Brewster angle for air-glass interface, when the refractive index of glass = 1.5
Answer:
According to Brewster law
tan i_B = μ
i_B = tan^-1 (μ)
i_B = tan^-1(l. 5)
i_B = 56.30

Question 12.
A polaroid (I) is placed in front of a monochromatic source. Another polaroid (II) is placed in front of this polaroid (I) and rotated till no light passes. A third polaroid (III) is now placed in between (I) and (II). In this case, will light emerge from (II). Explain. (NCERT Exemplar)
Answer:
Only in the special cases when the pass axis of (III) is parallel to (I) or (II), there shall be no light emerging. In all other cases, there shall be light emerging because the pass axis of (I) is no longer perpendicular to the pass axis of (III).

Question 13.
What is the shape of the wavefront of earth for sunlight? (NCERT Exemplar)
Answer:
Spherical with huge radius as compared to the earth’s radius so that it is almost a plane.

Question 14.
Consider a point at the focal point of a convergent lens. Another convergent lens of short focal length is placed on the other side. What is the nature of the wavefronts emerging from the final image? (NCERT Exemplar)
Answer:
The focal point of a convergent lens is the position of real image formed by this lens when object is at infinity. When another convergent lens of short focal length is placed on the other side, the combination will form a real point image at the combined focus of the two lenses. The wavefronts emerging from the final image will be spherical.

Short answer type questions

Question 1.
State two conditions required for obtaining coherent sources. In Young’s arrangement to produce interference pattern, show that dark and bright fringes appearing on the screen are equally spaced.
Answer:
Conditions for obtaining coherent sources:
(i) Coherent sources of light should be obtained from a single source by same device.
(ii) The two sources should give monochromatic light.
The separation between the centres or two consecutive bright fringes is the width of a dark fringe.

Hence, all bright and dark fringes are equally spaced on screen.

Question 2.
How will the interference pattern in Young’s double-slit experiment get affected, when
(i) distance between the slits S₁ and S₂ reduced and
(ii) the entire set-up is immersed in water? Justify your answer in each case.
Answer:
(i) The fringe width of interference pattern increases with the decrease in separation between S₁S₂ as
β ∝ \(\frac{1}{d}\)
(ii) The fringe width decrease as wavelength gets reduced when interference set up is taken from air to water.

Question 3.
What is the minimum angular separation between two stars, if a telescope is used to observe them with an objective of aperture 0.2 m? The wavelength of light used is 5900 A.
Answer:
Here, D = diameter of the objective of telescope = 0.2 m
λ = Wavelength of light used = 5900 Å = 5900 x 10^-10 m
Let dθ = Minimum angular separation between two stars =?
Using the relation,
dθ = \(\frac{1.22 \lambda}{D}\) , we get
dθ = \(\frac{1.22 \times 5900 \times 10^{-10}}{0.2}= \) = 3.6 x 10^-6 rad.

Question 4.
Distinguish between polarised and unpolarised light. Does the intensity of polarised light emitted by a polaroid depend on its orientation? Explain briefly. The vibrations in a beam of polarised light make an angle of 60° with the axis of the polaroid sheet. What percentage of light is transmitted through the sheet?
Answer:
A light which has vibrations in all directions in a plane perpendicular to the direction of propagation is said to be unpolarised light. The light from the sun, an incandescent bulb or a candle is unpolarised. If the electric field vector of a light wave vibrates just in one direction perpendicular to the direction of wave propagation, then it is said to be polarised or linearly polarised light.

Yes, the intensity of polarised light emitted by a polaroid depends on orientation of polaroid. When polarised light is incident on a polaroid, the resultant intensity of transmitted light varies directly as the square of the cosine of the angle between polarisation direction of light and the axis of the polaroid.

I ∝ cos² θ or I = I₀ cos² θ
where I₀ = maximum intensity of transmitted light;
θ = angle between vibrations in light and axis of polaroid sheet.
or I =I₀ cos² 60° = \(\frac{I_{0}}{4}\)
Percentage of light transmitted = \(\frac{I}{I_{0}} \) x 100 = \(\frac{1}{4}\) x 100 = 25%

Question 5.
Find an expression for intensity of transmitted light, when a polaroid sheet is rotated between two crossed polaroids. In which position of the polaroid sheet will the transmitted intensity be maximum?
Answer:
Let us consider two crossed polarizers, P₁ and P₂ with a polaroid sheet P₃ placed between them.

Let I₀ be the intensity of polarised light after passing through the first polarizer P₁.
If θ is the angle between the axes of P₁ and P₃, then the intensity of the polarised light after passing through P₃ will be I =I₀ cos²θ.
As P₁ and P₂ are crossed, the angle between the axes of P₁ and P₂ is 90°.
∴ The angle between the axes of P₂ and P₃ is (90° – 0).
The intensity of light emerging from P₂ will be given by

The intensity of polarised light transmitted from P₂ will be maximum, when ,
sin 2θ = maximum = 1
⇒ sin2θ = sin9O°
⇒ 2θ = 90°
⇒ θ = 45°
Also, the maximum transmitted intensity will be given by I = \(\frac{I_{0}}{4}\)

Question 6.
State Brewster’s law. The value of Brewster angle for a transparent medium is different for light of different colours. Give reason.
Answer:
Brewster’s Law: When unpolarized light is incident on the surface separating two media at polarising angle, the reflected light gets completely polarised only when the reflected light and the refracted light are perpendicular to each other. Now, refractive index of denser (second) medium with respect to rarer (first) medium is given by μ = tan i_B, where i_B = polarising angle.
Since refractive index is different for different colours (wavelengths), Brewster’s angle is different for different colours.

Question 7.
Can reflection result in plane polarised light if the light is incident on the interface from the side with higher refractive index? (NCERT Exemplar)
Answer:
When angle of incidence is equal to Brewster’s angle, the transmitted light is unpolarised and reflected light is plane polarised.

Consider the diagram in which unpolarised light is represented by dot and plane polarised light is represented by arrows.
Polarisation by reflection occurs when the angle of incidence is the Brewster’s angle
i.e., tani_B = ¹μ₂ = \(\frac{\mu_{2}}{\mu_{1}}\) where μ² < μ¹
when the light rays travels in such a medium, the critical angle is
sin i_c = \(\frac{\mu_{2}}{\mu_{1}}\)
where, μ₂ < μ₁
As | tani_B| > | sin i_C| for large angles i_B <i_C.
Thus, the polarisation by reflection occurs definitely.

Question 8.
Consider a two-slit interference arrangements (figure) such that the distance of the screen from the slits is half the distance between the slits. Obtain the value of D in terms of λ such that the first minima on the screen falls at a distance D from the centre O. (NCERT Exemplar)

Answer:
From the given figure of two-slit interference arrangements, we can write

The minima will occur when S₂P – S₁P = (2 n -1)\(\frac{\lambda}{2}\)
i.e., [D² +(D + X)²]^1/2 -[D² + (D -x)²]^1/2
= \(\frac{\lambda}{2}\)
[for first minima n = 1]
If x = D
We can write [D² +4D²]^1/2 -[D² +0]^1/2 = \(\frac{\lambda}{2}\)
⇒ [5D²]^1/2 – [D²]^1/2 = \(\frac{\lambda}{2}\)
⇒ \(\sqrt{5}\)D – D = \(\frac{\lambda}{2}\)

Long answer type questions

Question 1.
In Young’s double-slit experiment, deduce the conditions for (i) constructive, and (ii) destructive interference at a point on the screen. Draw a graph showing variation of the resultant intensity in the interference pattern against position ‘X’ on the screen.
Answer:
Conditions for Constructive and Destructive Interference :
When two waves of same frequency and constant initial phase difference travel in the same direction along a straight line simultaneously, they superpose in such a way that the intensity of the resultant wave is maximum at certain points and minimum at certain other points. The phenomenon of redistribution of intensity due to superposition of two waves of same frequency and constant initial phase difference is called the interference.

The waves of same frequency and constant initial phase difference are called coherent waves. At points of medium where the waves arrive in the same phase, the resultant intensity is maximum and the interference at these points is said to be constructive. On the other hand, at points of medium where the waves arrive in opposite phase, the resultant intensity is minimum and the interference at these points is said to be destructive. The positions of maximum intensity are called maxima while those of minimum intensity are called minima. The interference takes place in sound and light both.

Mathematical Analysis: Suppose two coherent waves travel in the same direction along a straight line, the frequency of each wave is \(\frac{\omega}{2 \pi}\) and amplitudes of electric field are a₁ and a₂ respectively. If at any time t, the electric fields of waves at a point are y₁ and y₂ respectively and phase difference is, Φ then equation of waves may be expressed as
y₁ = a₁ sin ωt ………………………. (1)
y₂ = a₂ sin ωt +Φ) ……………………………………….. (2)
According to Young’s principle of superposition, the resultant displacement at that point will be
y = y₁+y₂ ……………………………….. (3)
Substituting values of y₁ and y₂ from (1) and (2) in (3), we get
y = a₁ sin ωt + a₂ sin(ωt + Φ)

Using trigonometric relation,
sin(ωt +Φ) = sinωtcosΦ +cosωtsinΦ
y = a₁ sin ωt + a₂(sinωtcosΦ) + cosωt sin Φ)
= (a₁ +a₂cos Φ) sinωt + (a₂ sinΦ)cosωt …………………………….. (4)
Let a₁ + a₂ cosΦ = A cos θ ……………………………………… (5)
and a₂ sinΦ = A sinθ ………………………………………… (6)

Where A and θ are new constants
Then equation (4) gives
y = A cosθ sinωt + A sinθ cosωt
= A sin (ωt +θ) ……………………………………………. (7)
This is the equation of the resultant disturbance. Clearly the amplitude of resultants disturbance is A and phase difference from first wave is 0. The values of A and 0 are determined by (5) and (6). Squaring (5) and (6) and then adding, we get

∴ Amplitude,
A = \(\sqrt{a_{1}^{2}+a_{2}^{2}+2 a_{1} a_{2} \cos \phi} \) …………………………… (8)
As the intensity of a wave is proportional to its amplitude in arbitrary units I = A²
∴ Intensity of resultant wave,
I = A² = a₁² + a₂² + 2a₁a₂ cosΦ ……………………….. (9)
Clearly, the intensity of the resultant wave at any point depends on the amplitudes of individual waves and the phase difference between the waves at the point.

Constructive Interference: For maximum intensity at any point
cos Φ = +1
or phase difference Φ = 0,2π,4π,6π,……………………….
= 2nπ (n=0,1,2,3,……………………) …………………………………… (10)
The maximum intensity
I_max = a¹₂+a_2²
= (a₁+a₂)² …………………………..(11)
Path difference
Δ = \(\frac{\lambda}{2 \pi}\) x phase difference
= \(\frac{\lambda}{2 \pi} \) x 2nπ …………………………………………. (12)
Clearly, the maximum intensity is obtained in the region of superposition at those points where waves meet in the same phase or the phase difference between the waves is even multiple of π or path difference between them is the integral multiple of λ and maximum intensity is (a₁ +a₂)²

which is greater than the sum intensities of individual waves by an amount 2a₁a₂.
Destructive Interference : For minimum intensity at any point CosΦ = -1
or phase difference,
Φ = π,3π,5π,7π, …………………………..
– (2n-l)π, n = 1,2,3,… …………………………………. (13)
In this case the minimum intensity,
I_min =a₁² +a₂² – 2a₁a₂
= (a₁-a₂)² ………………………… (14)

Path difference, Δ = \(\frac{\lambda}{2 \pi}\) x Phase difference
= \(\frac{\lambda}{2 \pi}\) x (2n – 1)π
= (2n-l) \(\frac{\lambda}{2}\)

Clearly, the minimum intensity is obtained in the region of superposition at those points where waves meet in opposite phase or the phase difference between the waves is odd multiple of π or path difference between the waves is odd multiple of \(\frac{\lambda}{2}\) and minimum intensity = (a₁ -a₂)² which is less than the sum of intensities of the individual waves by an amount 2a₁a₂.

From equations (12) and (14) it is clear that the intensity 2a₁a₂ is transferred from positions of minima to maxima, this implies that the interference is based on conservation of energy i.e., there is no wastage of energy.
Variation of Intensity of light with position x is shown in fig.

Question 2.
Describe diffraction of light due to a single slit. Explain formation of a pattern of fringes obtained on the screen and plot showing variation of intensity with angle θ in single slit diffraction.
Answer:
Diffraction of Light at a Single Slit: When monochromatic light is made incident on a single slit, we get diffraction pattern on a screen placed behind the slit. The diffraction pattern contains bright and dark bands, the intensity of central band is maximum and goes on decreasing on both sides.

Explanation: Let AB be a slit of width ‘a’ and a parallel beam of monochromatic light is incident on it. According to Fresnel, the diffraction pattern is the result of superposition of a large number of waves, starting from different points of illuminated slit.

Let θ be the angle of diffraction for waves reaching at point P of screen and AN the perpendicular dropped from A on wave diffracted from B. The path difference between rays diffracted at points A and B,
Δ = BP – AP = BN
In ΔANB, ∠ANB = 90°
and ∠BAN = θ
∴ sinθ = \(\frac{B N}{A B}\) or BN = AB sinθ
As AB = width of slit = a
Path difference Δ = asinθ ……………………………… (1)

To find the effect of all coherent waves at P, we have to sum up their contribution, each with a different phase. This was done by Fresnel by rigorous calculations, but the main features may be explained by simple arguments given below :
At the central point C of the screen, the angle 0 is zero. Hence the waves starting from all points of slit arrive in the same phase. This gives maximum intensity at the central point C. If point P on screen is such that the path difference between rays starting from edges A and B is λ, then path difference,

If angle θ is small,
sinθ = θ = \(\frac{\lambda}{a}\) ……………………………. (2)
Minima: Now we divide the slit into two equal halves AO and OB, each of width \(\frac{a}{2}\).
Now for every point, M₁ in AO, there is a corresponding point M₂ in OB, such that M₁M₂ = \(\frac{a}{2}\) .
Then path difference between waves arriving at P and starting from M₁ and M₂ will be \(\frac{a}{2}\) sin θ = \(\frac{\lambda}{2}\).

This means that the contributions from the two halves of slit AO and OB are opposite in phase and so cancel each other. Thus equation (2) gives the angle of diffraction at which intensity falls to zero. Similarly it may be shown that the intensity is zero for sin θ = \(\frac{n \lambda}{a}\) , with n as integer. Thus, the general condition of minima is asinθ = nλ ……………………………………… (3)

Secondary Maxima: Let us now consider angle θ such that
sin θ = θ = \(\frac{3 \lambda}{2 a}\)

Which is midway between two dark bands given by
sin θ = θ = \(\frac{\lambda}{a} \) and sin θ = θ = \(\frac{2 \lambda}{a}\)
Let us now divide the slit into three parts. If we take the first two parts of slit, the path difference between rays diffracted from the extreme ends of the first two parts.
\(\frac{2}{3}\) a sin θ = \(\frac{2}{3} a \times \frac{3 \lambda}{2 a}\) = λ

Then the first two parts will have a path difference of \(\frac{\lambda}{2}\) and cancel the effect of each other. The remaining third part will contribute to the intensity at a point between two minima. Clearly, there will be maxima between first two minima, but this maximum will be of much weaker intensity than central maximum.

This is called first secondary maxima. In a similar manner, we can show that there are secondary maxima between any two consecutive minima; and the intensity of maxima will go on decreasing with increase of order of maxima.
In general, the position of nth maxima will be given by
a sin θ = \(\left(n+\frac{1}{2}\right)\) λ (n =1, 2, 3, 4,…) ………………………………… (4)
The intensity of secondary maxima decreases with increase of order n because with increasing n, the contribution of slit decreases.
For n = 2, it is one-fifth, for n = 3, it is one-seventh and so on.

Punjab State Board PSEB 12th Class Physics Important Questions Chapter 5 Magnetism and Matter Important Questions and Answers.

PSEB 12th Class Physics Important Questions Chapter 5 Magnetism and Matter

Very short answer type questions

Question 1.
What is the angle of dip at a place where the horizontal and vertical components of the earth’s magnetic field are equal?
Answer:
The angle of dip is given by
θ = tan^-1 (\(\frac{B_{V}}{B_{H}}\))
B_V = vertical component of the earth’s magnetic field.
B_H = horizontal component of the earth’s magnetic field.
So, as B_V = B_H
Then, θ = tan^-1 (1) = 45°
∴ The angle of dip will be θ = 45°.

Question 2.
The motion of copper plate is damped when it is allowed to oscillate between the two poles of a magnet. What is the cause of this damping?
Answer:
As the copper plates oscillate in the magnetic field between the two plates of the magnet, there is a continuous change of magnetic flux linked with the pendulum. Due to this, eddy currents are set up in the copper plate which try to oppose the motion of the pendulum according to the Lenz’s law and finally bring it to rest.

Question 3.
Relative permeability of a material μ_r = 0.5. Identify the nature of the magnetic material and write its relation of magnetic susceptibility.
Answer:
The nature of magnetic material is a diamagnetic.
μ_r = 1 + χ_m

Question 4.
Which of the following substances are diamagnetic?
Bi, Al, Na, Cu, Ca and Ni
Answer:
Diamagnetic substances are (i) Bi (ii) Cu.

Question 5.
The susceptibility of a magnetic material is -4.2 × 10^-6. Name the type of magnetic material, it represents.
Answer:
Negative susceptibility represents diamagnetic substance.

Question 6.
What are permanent magnets? Give one example.
Answer:
Substances that retain their attractive property for a long period of time at room temperature are called permanent magnets.
Examples: Those pieces which are made up of steel, alnico, cobalt and ticonal.

Question 7.
Why is the core of an electromagnet made of ferromagnetic materials?
Answer:
Ferromagnetic material has a high retentivity. So on passing current through windings it gains sufficient magnetism immediately.

Question 8.
From molecular view point, discuss the temperature dependence of susceptibility for diamagnetism, paramagnetism and ferromagnetism. (NCERT Exemplar)
Answer:
Diamagnetism is due to orbital motion of electrons developing magnetic moments opposite to applied field and hence is not much affected by temperature.

Paramagnetism and ferromagnetism is due to alignments of atomic magnetic moments in the direction of the applied field. As temperature increases, this alignment is disturbed and hence susceptibilities of both decrease as temperature increases.

Question 9.
A ball of superconducting material is dipped in liquid nitrogen and placed near a bar magnet.
(i) In which direction will it move?
(ii) What will be the direction of its magnetic moment? (NCERT Exemplar)
Answer:
When a diamagnetic material is dipped in liquid nitrogen, it again behaves as a diamagnetic material. Thus, superconducting material will again behave as a diamagnetic material. When this diamagnetic material is placed near a bar magnet, it will be feebly magnetised opposite to the direction of magnetising field.

(i) Thus, it will be repelled.
(ii) Also its direction of magnetic moment will be opposite to the direction of magnetic field of magnet.

Question 10.
Consider the plane S formed by the dipole axis and the axis of earth. Let P be point on the magnetic equator and in S. Let Q be the point of intersection of the geographical and magnetic equators. Obtain the declination and dip angles at P and Q. (NCERT Exemplar)
Answer:
In adjoining figure:

(i) P is in S (needle will point both north)
Declination = 0
P is also on magnetic equator.
∴ Dip = 0

(ii) Q is on magnetic equator.
∴ Dip = 0
But declination = 11.3.

Short answer type questions

Question 1.
Explain the following:
(i) Why do magnetic field lines form continuous closed loops?
(ii) Why are the field lines repelled (expelled) when a
diamagnetic material is placed in an external uniform magnetic field?
Answer:

(i) Magnetic lines of force form continuous closed loops because a magnet is always a dipole and as a result, the net magnetic flux of a magnet is always zero.
(ii) When a diamagnetic substance is placed in an external magnetic field, a feeble magnetism is induced in opposite direction. So, magnetic lines of force are repelled.

Question 2.
How does a circular loop carrying current behaves as a magnet?
Answer:
The current round in the face of the coil is in anti-clockwise direction, then this behaves like a North pole, whereas when it viewed from other scale, then current round in it is in clockwise direction necessarily forming South pole of magnet.

Hence, current loop have both magnetic poles and therefore, behaves like a magnetic dipole.

Question 3.
Give two points to distinguish between a paramagnetic and diamagnetic substance.
Answer:

Paramagnetic substance	Diamagnetic substance
1. A paramagnetic substance is feebly attracted by magnet.	A diamagnetic substance is feebly repelled by a magnet.
2. For a paramagnetic substance, the intensity of magnetisation has a small positive value.	For a diamagnetic substance, the intensity of magnetism has a small negative value.

Question 4.
(a) How is an electromagnet different from a permanent magnet?
Write two properties of a material which makes it suitable for making (i) a permanent magnet, and (ii) an electromagnet.
Answer:
(a) An electromagnet consists of a core made of a ferromagnetic material placed inside a solenoid. It behaves like a strong magnet when current flows through the solenoid and effectively loses its magnetism when the current is switched off.

A permanent magnet is also made up of a ferromagnetic material but it retains its magnetism at room temperature for a long time after being magnetised one.

(b) Properties of material are as below:
(i) Permanent magnet

• Retentivity and coercivity should be large
• Magnetically hard

(ii) An electromagnet

• Magnetically soft
• Coercivity should be low.

Question 5.
A bar magnet of magnetic moment M and moment of inertia I (about centre, perpendicular to length) is cut into two equal pieces, perpendicular to length. Let T be the period of oscifiations of the original magnet about an axis through the mid-point, perpendicular to length, in a magnetic field B. What would be the similar period T’ for each piece? (NCERT Exemplar)
Answer:
Given, I = moment of inertia of the bar magnet
m = mass of bar magnet
l = length of magnet about an any passing through its centre and perpendicular to its length
M = magnetic moment of the magnet
B = uniform magnetic field in which magnet is oscillating, we get time period of oscillation is
T = 2π\(\sqrt{\frac{I}{M B}}\)
Here I = \(\frac{m l^{2}}{12}\)
When magnet is cut into two equal pieces, perpendicular to length, then moment of inertia of each piece of magnet about an axis perpendicular to length passing through its centre is

Question 6.
A uniform conducting wire of length 12 a and resistance R is wound up as a current carrying coil in the shape of (i) an equilateral triangle of side a; (ii) a square of sides a and, (iii) a regular hexagon of sides a. The coil is connected to a voltage source V₀. Find the magnetic moment of the coils in each case. (NCERT Exemplar)

(i) Area of equilateral triangle, A = \(\frac{\sqrt{3}}{4}\) a²
(ii) Area of square, A = a²

Long answer type questions

Question 1.
Verify the Ampere’s law for magnetic field of a point dipole of dipole moment M = Mk̂. Take C as the closed curve running clockwise along
(i) the z-axis from z a > 0 to z = R,
(ii) along the quarter circle of radius R and centre at the origin in the first quadrant of vz-plane,
(iii) along the x-axis from x = R to x = a, and
(iv) along the quarter circle of radius a and centre at the origin
in the first quadrant of xz-plane (NCERT Exemplar)
Answer:
From P to Q, every point on the z-axis lies at the axial line of magnetic
dipole of moment \(\vec{M}\). Magnetic field induction at a point distance z from the magnetic dipole of moment is

(ii) Along the quarter circle QS of radius R as-.given in the figure below

The point A lies on the equatorial line of the magnetic dipole of moment M sin0. Magnetic field at point A on the circular arc is
B = \(\frac{\mu_{0}}{4 \pi} \frac{M \sin \theta}{R^{3}}\) ; \(\overrightarrow{d l}\) = Rdθ

(iii) Along x-axis over the path ST, consider the figure given below.

From figure, every point lies on the equatorial line of magnetic dipole. Magnetic field induction at a point distance x from the dipole is

(iv) Along the quarter circle TP of radius a. Consider the figure given below

Punjab State Board PSEB 12th Class Physics Important Questions Chapter 11 Dual Nature of Radiation and Matter Important Questions and Answers.

PSEB 12th Class Physics Important Questions Chapter 11 Dual Nature of Radiation and Matter

Very short answer type questions

Question 1.
In the photoelectric effect, why should the photoelectric current increase as the intensity of monochromatic radiation incident on a photosensitive surface is increased? Explain.
Answer:
The photoelectric current increases proportionally with the increase in intensity of incident radiation. Larger the intensity of incident radiation, larger is the number of incident photons and hence larger is the number of electrons ejected from the photosensitive surface.

Question 2.
Define the term ‘threshold frequency’ in relation to photoelectric effect.
Answer:
Threshold frequency is defined as the minimum frequency of incident radiation which can cause photoelectric emission. It is different for different metal.

Question 3.
WrIte the basic features of photon picture of electromagnetic radiation on which Einstein’s photoelectric equation is based.
Answer:
Features of the photons are as:

• Photons are particles of light having energy E = hv and momentum p = \(\frac{h}{\lambda}\), where h is Planck’s constant.
• Photons travel with the speed of light in vacuum, independent of the frame of reference.
• Intensity of light depends on the number of photons crossing unit area in a unit time.

Question 4.
Define Intensity of radiation on the basis of photon picture of light. Write its SI unit.
Answer:
The amount of light energy or photon energy incident per meter square per second is called intensity of radiation1 Its SI unit is \(\frac{\mathrm{W}}{\mathrm{m}^{2}}\) or J/s m

Question 5.
State de Broglie hypothesis.
Answer:
According to the hypothesis of de Brogue “The atomic particles of matter moving with a given velocity, can display the wave-like properties.” i.e., λ = \(\frac{h}{m v}\) (mathematically)

Question 6.
Write the relationship of de Brogue wavelength λ associated with a particle of mass m terms of its kinetic energy E.
Answer:
Kinetic energy E_K = \(\frac{p^{2}}{2 m}\)
where, p = momentum
m = mass and E_K = kinetic energy
⇒ p = \(\sqrt{2 m E_{K}} \)
de Brogue wavelength,
λ = \(\frac{h}{p}\)
Where, p = \(\sqrt{2 m E_{K}} \)
⇒ λ = \(\frac{h}{\sqrt{2 m E_{K}}}\)

Question 7.
Name the phenomenon which shows the quantum nature of electromagnetic radiation.
Answer:
Photoelectric effect.

Question 8.
Do all the electrons that absorb a photon comes out as photoelectrons? (NCERT Exemplar)
Answer:
No, most electrons get scattered into the metal. Only a few come out of the surface of the metal.

Question 9.
There are materials which absorb photons of shorter wavelength and emit photons of longer wavelength. Can there be stable substances which absorb photons of larger wavelengths and emit light of shorter wavelengths? (NCERT Exemplar)
Answer:
In the first case, energy given out is less than the energy supplied. In the second case, the material has to supply the energy as the emitted photon has more energy. This cannot happen for stable substances.

Question 10.
There are two sources of light, each emitting with a power 100W.
One emits X-rays of wavelength 1 nm and the other visible light at 500 nm.
Find the ratio of number of photons of X-rays the photons of visible light of the given wavelength. (NCERT Exemplar) Ans. Total E is constant.
Let n₁ and n₂ be the number of photons of X-rays and visible region.
n₁E₂ = n₂E₂

Short answer type questions

Question 1.
What is meant by work function of a metal? How does the value of work function influence the kinetic energy of electrons liberated during photoelectron emission?
Answer:
Work Function: The minimum energy required to free an electron from metallic surface is called the work function.
Smaller the work function, larger the kinetic energy of emitted electron.

Question 2.
Show mathematically how Bohr’s postulate of quantization of orbital angular momentum in hydrogen atom is explained by de Broglie’s hypothesis.
Answer:
According to de Broglie’s hypothesis,
λ = \(\frac{h}{m v}\) …………………………… (1)
According to de Broglie’s condition of stationary orbits, the stationary orbits are those which contain complete de Broglie wavelength.
2πr = nλ …………………………………… (2)

Substituting value of λ from (1) in (2), we get
2πr = n\(\frac{h}{m v}\)
⇒ mvr = n\(\frac{h}{2 \pi}\) ……………………………………… (3)
This is Bohr’s postulate of quantization of energy levels.

Question 3.
Write briefly the underlying principle used in the Davison-Germer experiment to verify wave nature of electrons experimentally. What is the de Broglie wavelength of an electron with kinetic energy (E_K) 120 eV?
Answer:
Principle: Diffraction effects are observed for beams of electrons scattered by the crystals.
λ = \(\frac{h}{p}=\frac{h}{\sqrt{2 m E_{K}}}=\frac{h}{\sqrt{2 m e V}}\)
= \(\frac{6.63 \times 10^{-34}}{\sqrt{2 \times 9.1 \times 10^{-31} \times 1.6 \times 10^{-19} \times 120}}\)
λ = 0.112 nm.

Question 4.
(i) Describe briefly three experimentally observed features in the phenomenon of photoelectric effect.
(ii) Discuss briefly how wave theory of light cannot explain these features.
Answer:
(i) Three experimentally observed features in the phenomenon of photoelectric effect are as follows :

• Intensity: When intensity of incident light increases as one photon ejects one electron, the increase in intensity will increase the number of ejected electrons. Frequency has no effect on photoelectron.
• Frequency: When the frequency of incident photon increases, the kinetic energy of the emitted electrons increases. Intensity has no effect on kinetic energy of photoelectrons.
• No Time Lag: When energy incident photon is greater than the work function, the photoelectron is immediately ejected. Thus, there is no time lag between the incidence of light and emission of photoelectrons.

(ii) These features cannot be explained in the wave theory of light because wave nature of radiation cannot explain the following :

• The instantaneous ejection of the photoelectrons.
• The existence of threshold frequency for a metal surface.
• The fact that kinetic energy of the emitted electrons is independent of the intensity of light and depends upon its frequency.

Question 5.
A beam of monochromatic radiation is incident on a photosensitive surface. Answer the following questions :
(i) Do the emitted photoelectrons have the same kinetic energy?
(ii) Does the kinetic energy of the emitted electrons depend on the intensity of incident radiation?
(iii) On what factors does the number of emitted photoelectrons depend?
Answer:
In photoelectric effect, an electron absorbs a quantum of energy hv of radiation, which exceeds the work function, an electron is emitted with maximum kinetic energy.
E_{K max} = hv – W
(i) No, all electrons are bound with different forces in different layers of the metal. So, more tightly bound electron will emerge with less kinetic energy. Hence, all electrons do not have same kinetic energy.
(ii) No, because an electron cannot emit out if quantum energy hv is less than the work function of the metal. The K_E depends on the energy of each photon.
(iii) Number of emitted photoelectrons depends on the intensity of the radiations provided the quantum energy hv is greater than the work function of the metal.

Question 6.
Define the term “cut-off frequency” in photoelectric emission. The threshold frequency of a metal is f. When the light of frequency 2f is incident on the metal plate, the maximum velocity of photoelectrons is v₁. When the frequency of the incident radiation is increased to 5f, the maximum velocity of photoelectrons is v₂. Find the ratio v₁: v₂.
Answer:
Cut-off Frequency: It is that maximum frequency of incident radiation below which no photoemission takes place from a photoelectric material. According to Einstein’s photoelectric equation
E_{K max} = \(\frac{h c}{\lambda}-\phi \) = \(h v-\phi\)
Given that threshold frequency of the metal is f. If light of frequency, 2f is incident on metal plate, maximum velocity of photoelectron is v₁ then,
\(\frac{1}{2} m v_{1}^{2}\) = h (2f-f)
⇒ \(\frac{1}{2} m v_{1}^{2}\) = hf …………………….. (1)

If light of frequency, 5f is incident and maximum velocity of photoelectron is v₂.
\(\frac{1}{2} m v_{1}^{2}\) = h(5f-f)
⇒ \(\frac{1}{2} m v_{1}^{2}\) = 4hf ………………………………… (2)
Dividing (1) by (2), we get
\(\left(\frac{v_{1}}{v_{2}}\right)^{2}=\frac{1}{4}\)
⇒ \(\frac{v_{1}}{v_{2}}=\frac{1}{2}\)
∴ v₁ = v₂ = 1:2

Question 7.
Two monochromatic beams A and B of equal intensity I hit a screen. The number of photons hitting the screen by beam A is twice that by beam B. Then what inference can you make about their frequencies? (NCERTExemplarl
Answer:
Let no. of photons falling per second of beam A = n_A
No. of photons falling per second of beam B = n_B
Energy of beam A = hv_A
Energy of beam B = h v_B

According to question, I = n_Av_A = n_Bv_B
\(\frac{n_{A}}{n_{B}}=\frac{v_{B}}{v_{A}}\) or \(\frac{2 n_{B}}{n_{B}}=\frac{v_{B}}{v_{A}}\)
⇒ v_B = 2 v_A
The frequency of beam B is twice that of A.

Question 8.
Consider Fig. for photoemission.

How would you reconcile with momentum- conservation? No light (Photons) have momentum in a different direction than the emitted electrons. (NCERT Exemplar)
Answer:
The momentum is transferred to the metal. At the microscopic level, atoms absorb the photon and its momentum is transferred mainly to the nucleus and electrons. The excited electron is emitted. Conservation of momentum needs to be accounted for the momentum transferred to the nucleus and electrons.

Long answer type questions

Question 1.
Describe Davisson and Germer’s experiment to demonstrate the wave nature of electrons. Draw a labeled diagram of apparatus used.
Answer:
Davisson and Germer Experiment: In 1927 Davisson and Germer performed a diffraction experiment with electron beam in analogy with X-ray diffraction to observe the wave nature of matter.
Apparatus: It consists of three parts
(i) Electron gun : It gives a fine beam of electrons, de Brogue used electron beam of energy 54 eV. de Brogue wavelength associated with this beam

λ = \(\frac{h}{\sqrt{2 m E_{k}}}\)
Here, m = mass of electron = 9.1 x 10^-31 kg
E_K = Kinetic energy of electron = 54eV
= 54 x 1.6 x 10^-19 J = 86.4 x 10^-19 J
∴ λ = \(\frac{6.63 \times 10^{-34}}{\sqrt{2 \times 9.1 \times 10^{-31} \times 86.4 \times 10^{-19}}}\)
= 1.66 x 10^-10 = 1.66 Å
(ii) Nickel crystal: The electron beam was directed on nickel crystal against the electron detector. The smallest separation between nickel atoms is O.914Å. Nickel crystal behaves as diffraction grating.

(iii) Electron detector: It measures the intensity of electron beam diffracted from nickel crystal. It may be an ionization chamber fitted with a sensitive galvanometer. The energy of electron beam, the angle of incidence of beam on nickel crystal and the position of detector can all be varied.

Method: The crystal is rotated in small steps to change the angle (α say) between incidence and scattered directions and the corresponding intensity (I) of scattered beam is measured. The variation of the intensity (I) of the scattered electrons with the angle of scattering a is obtained for different accelerating voltages.

The experiment was performed by varying the accelerating voltage from 44 V to 68 V. k was noticed that a strong peak appeared in the intensity (I) of the scattered electron for an accelerating voltage of 54 V at a scattering angle α = 50°.

From Bragg’s law, 2d sinθ = nλ
Here, n =1,d =0.914 Å,θ =65°
∴ λ = \(\frac{2 d \sin \theta}{n}=\frac{2 \times(0.914 \AA) \sin 65^{\circ}}{1}\)
=2 x 0.914 x 0.9063Å =1.65Å
The measured wavelength is in close agreement with the estimated de Broglie wavelength. Thus the wave nature of electrons is verified. Later on G.P. Thomson demonstrated the wave nature of fast electrons. Due to their work Davisson and G.P. Thomson were awarded Nobel Prize in 1937.

Punjab State Board PSEB 12th Class Physics Important Questions Chapter 4 Moving Charges and Magnetism Important Questions and Answers.

PSEB 12th Class Physics Important Questions Chapter 4 Moving Charges and Magnetism

Very short answer type questions

Question 1.
Using the concept of force between two infinitely long parallel current carrying conductors define one ampere of current.
Answer:
One ampere is that value of current which flows through two straight, parallel infinitely long current carrying conductors placed in air or. vacuum at a distance of 1 m and they experience a force of attractive or repulsive nature of magnitude 2 × 10^-7 N/m on their unit length.

Question 2.
State Ampere’s circuit law.
Answer:
It states that the line integral of the magnetic field \(\vec{B}\) around any closed circuit is equal to p0 times the total current passing through this closed circuit.
\(\oint \vec{B} \cdot \overrightarrow{d l}\) = μ₀ I

Question 3.
A narrow beam of protons and deuterons, each having the same momentum, enters a region of uniform magnetic field directed perpendicular to their direction of momentum. What would be the ratio of the radii of the circular path described by them?
Answer:
For the given momentum of charge particle, radius of circular paths depends on charge and magnetic field as
r = \(\) ⇒ r ∝ \(\)
For given momentum,
∴ r_proton : r_deuteron = 1 : 1
As they have same momentum and charge moving in a small magnetic field.

Question 4.
Write the expression, in a vector form, for the Lorentz magnetic force \(\overrightarrow{\boldsymbol{F}}\) due to a charge moving with velocity \(\vec{v}\) in a magnetic field \(\overrightarrow{\boldsymbol{B}}\). What is the direction of the magnetic force?
Answer:
Force, \(\vec{F}=q(\vec{v} \times \vec{B})\)
Obviously, the force on charged particle is perpendicular to both velocity \(\vec{v}\) and magnetic field \(\vec{B}\).

Question 5.
When a charged particle moving with velocity \(\vec{v}\) is subjected to magnetic field \(\overrightarrow{\boldsymbol{B}}\), the force acting on it is non-zero. Would the particle gain any energy?
Answe:
No. (i) This is because the charge particle moves on a circular path.
(ii) \(\vec{F}=q(\vec{\nu} \times \vec{B})\)
and power dissipated p = \(\vec{F} \times \vec{V}\)
= q \((\vec{v} \times \vec{B}) \times \vec{y}\) = p\((\vec{v} \times \vec{v}) \times \vec{B}\)
The particle does not gain any energy.

Question 6.
A square coil OPQR of side a carrying a current 7, is placed in the Y-Z plane as shown here. Find the magnetic moment associated with this coil.

Answer:
The magnetic moment associated with the coil, is \(\vec{\mu}\)_m = Ia²î

Question 7.
Under what condition is the force acting on a charge moving through a uniform magnetic field is minimum?
Answer:
F_m = qvB sinθ; for minimum force sinθ = 0. i.e., force is minimum when charged particle move parallel or anti-parallel to the field.

Question 8.
What is the nature of magnetic field in a moving coil galvanometer?
Answer:
The nature of magnetic field in a moving coil galvanometer is radial.

Question 9.
Verify that the cyclotron frequency ω = eB/m has the correct dimensions of [T]^-1. (NCERT Exemplar)
Or A charged particle of charge e and mass m is moving in an electric field E and magnetic field B. Construct dimensionless quantities and quantities of dimension [T]^-1. (NCERT Exemplar)
Answer:
For a charge particle moving perpendicular to the magnetic field, the magnetic Lorentz forces provides necessary centripetal force for revolution.
\(\frac{m v^{2}}{R}\) = qvB
On simplifying the terms, we have
∴ \(\frac{q B}{m}=\frac{v}{R}\) = ω
Finding the dimensional formula of angu
∴ [ω] = \(\left[\frac{q B}{m}\right]=\left[\frac{v}{R}\right]\) = [T] ^-1

Question 10.
Show that a force that does no work must be a velocity dependent force. (NCERT Exemplar)
Answer:
Let no work is done by a force, so we have
dW = F.dl = 0
⇒ F. v dt = 0 (Since, dl = v dt and dt ≠ 0)
⇒ F.v = 0
Thus, F must be velocity dependent which implies that angle between F and v is 90°. If v changes (direction), then (directions) F should also change so that above condition is satisfied.

Question 11.
The magnetic force depends on v which depends on the inertial frame of reference. Does then the magnetic force differ from inertial frame to frame? Is it reasonable that the net acceleration has a different value in different frames of reference? (NCERT Exemplar|
Answer:
Yes, the magnetic force differ from inertial frame to frame. The magnetic force is frame dependent.
The net acceleration which comes into existing out of this is however, frame independent (non-relativistic physics) for inertial frames.

Question 12.
An electron enters with a velocity υ = υ₀î into a cubical region (faces parallel to coordinate planes) in which there are uniform electric and magnetic fields. The orbit of the electron is found to spiral down inside the cube in plane parallel to the x-y plane. Suggest a configuration of fields E and B that can lead to it. (NCERT Exemplar)
Answer:
Considering magnetic field B = B₀k̂, and an electron enters with a velocity v = v₀î into a cubical region (faces parallel to coordinate planes). The force on electron, using magnetic Lorentz force, is given by
F = -e(v₀î x B₀k̂) = ev₀B₀î
which revolves the electron in x-y plane.
The electric force F = -eE₀k̂ accelerates e along z-axis which in turn increases the radius of circular path and hence particle traversed on spiral path.

Short answer type questions

Question 1.
Write any two important points of similarities and differences each between Coulomb’s law for the electrostatic field and . Biot-Savart’s law for the magnetic field.
Answer:
Similarities: Both electrostatic field and magnetic field

• follows the principle of superposition.
• depends inversely on the square of distance from source to the point of interest.

Differences I

• Electrostatic field is produced by a scalar source (q) and the magnetic
  field is produced by a vector source (I\(\overrightarrow{d l}\)).
• Electrostatic field is along the displacement vector between source and point of interest; while magnetic field is perpendicular to the plane, containing the displacement vector and vector source.
• Electrostatic field is angle independent, while magnetic field is angle
  dependent between source vector and displacement vector.

Question 2.
State the underlying principle of a cyclotron. Write briefly how this machine is used to accelerate charged particles to high energies.
Or State the principle of the working of a cyclotron. Write two uses of this machine.
Answer:
The combination of crossed electric and magnetic fields is used to increase the energy of the charged particle. Cyclotron uses the fact that the frequency of revolution of the charged particle in a magnetic field is independent of its energy.

Inside the dees the particle is shielded from the electric field and magnetic field acts on the particle and makes it to go round in a circular path inside a dee.

Every time, the particle moves from one dee to the other it comes under the influence of electric field which ensures to increase the energy of the particle as the sign of the electric field changed alternately.
The increased energy increases the radius of the circular path so the accelerated particle moves in a spiral path.
Since, radius of trajectory
r = \(\frac{v m}{q B}\)
∴ v = \(\frac{r q B}{m}\)
Hence, the kinetic energy of ions
= \(\frac{1}{2}\)mv² = \frac{1}{2}\(\)m\(\frac{r^{2} q^{2} B^{2}}{m^{2}}\)
⇒ KE = \(\frac{1}{2}\)\(\frac{r^{2} q^{2} B^{2}}{m}\)

Question 3.
(i) State Ampere’s circuital law expressing it in the integral form.
(ii) Two long co-axial insulated solenoids S₁ and S₂ of equal length are wound one over the other as shown in the figure. A steady current I flows through the inner solenoid S₁ to the other end B which is connected to the outer solenoid S₂ through which the some current I flows in the opposite direction so, as to come out at end A. If n₁ and n₂ are the number of turns per unit length, find the magnitude and direction of the net magnetic field at a point
(a) inside on the axis and
(b) outside the combined system.

Answer:
(i) Ampere’s circuital law states that the line integral of magnetic field
(B) around any closed path in vacuum is μ₀ times the net current (I) threading the area enclosed by the curve.
Mathematically, \(\oint \vec{B} \cdot d \vec{l}\) = μ₀I
Ampere’s law is applicable only for an Amperian loop as the Gauss’s law is used for Gaussian surface in electrostatics.

(ii) According to Ampere’s circuital law, the net magnetic field is given by
B = μ₀nî.
(a) The net magnetic field is given by
B_net = B₂ – B₁
μ₀n₂I₂ – ₀n₁I₁
= μ₀I(n₂ – n₁)
The direction is from B to A.

(b) As the magnetic field due to Sx is confined solely inside S₁ as the solenoids are assumed to be very long. So, there is no magnetic field outside S₁ due to current in S₁, similarly there is no field outside S₂.
B_net = 0

Question 4.
(a) State Biot-Savart law and express this law in the vector
form.
(b) Two identical circular coils, P and Q each of radius R, carrying currents 1 A and √3 A, respectively, are placed concentrically and perpendicular to each other lying in the XY and YZ planes. Find the magnitude and direction of the net magnetic field at the centre of the coils. ’
Answer:
(a) According to Biot-Savart’s law the magnitude of the magnetic field \(\overrightarrow{d B}\) due to a small element of length dl of a current carrying wire at a point P, is proportional to the current I, the element length dl and is inversly proportional to the square of the distance r. It is also proportional to sinθ,

where θ is the angle between \(\overrightarrow{d l}\) and \(\vec{r}\).

Its direction is perpendicular to the plane containing \(\overrightarrow{d l}\) and \(\vec{r}\) in vector form
\(\overrightarrow{d B}\) ∝ \(\frac{I \overrightarrow{d l} \times \vec{r}}{r^{3}}\)
⇒ \(\overrightarrow{d B}\) = \(\frac{\mu_{0}}{4 \pi} \cdot \frac{I d \vec{l} \cdot \vec{r}}{r^{3}}\)
(\(\overrightarrow{d l}\) is directed along the length of the wire in the direction of current and
\(\vec{r}\) is the vector joining the centre of current element to the point P) (b) Field due to current in coil P is
\(\vec{B}\)₂ = \(\frac{\mu_{0} I_{1}}{2 R}\) .k̂
(Assuming current to be anticlockwise as seen form + ve Z-axis) and that due to current in coil Q is
\(\vec{B}\)₂ = \(\frac{\mu_{0} I_{2}}{2 R} \hat{i}\)
(Assuming current to be anticlockwise as seen form positive X-axis)

Question 5.
Explain, giving reasons, the basic difference in converting a galvanometer into (i) a voltmeter and (ii) an ammeter.
Answer:
(i) In converting a galvanometer into a voltmeter, a very high suitable resistance is connected in series to its coil. So, the galvanometer gives full scale deflection.
(ii) In converting a galvanometer into an ammeter, a very small suitable resistance is connected in parallel to its coil. The remaining pair of the current i.e., (I – I_g) flows through the resistance. Here I = Circuit current
and I_g = Current through galvanometer.

Question 6.
A current carrying loop consists of 3 identical quarter circles of radius R, lying in the positive quadrants of the x-y, y-z and z-x planes with their centres at the origin, joined together. Find the direction and magnitude of B at the origin. (NCERT Exemplar)
Answer:
For the current carrying loop quarter circles of radius R, lying in the positive quadrants of the x-y plane
B₁ = \(\frac{\mu_{0}}{4 \pi} \frac{I(\pi / 2)}{R}\) = k̂\(\frac{\mu_{0}}{4} \frac{I}{2 R}\) k̂
For the current carrying loop quarter circles of radius R, lying in the positive quadrants of the y-z plane
B₂ = \(\frac{\mu_{0}}{4} \frac{I}{2 R}\)î
For the current carrying loop quarter circles of radius R, lying in the positive quadrants of the z-x plane
B₃ = \(\frac{\mu_{0}}{4} \frac{I}{2 R}\)
Current carrying loop consists of 3 identical quarter circles of radius R, lying in the positive quadrants of the x-y, y-y and z-z planes with their centres at the origin, joined together is equal to the vector sum of magnetic field due to each quarter and given by,
B = \(\frac{1}{4 \pi}\) (î + ĵ + k̂)\(\frac{\mu_{0} I}{2 R}\)

Question 7.
A long straight wire carrying current of 25 A rests on a table as shown in figure. Another wire PQ of length 1 m, mass 2.5 g carries the same current but in the opposite direction.
The wire PQ is free to slide up and down. To what height will PQ rise? (NCERT Exemplar)

Answer:
The magnetic field produced by long straight wire carrying current of 25 A rests on a table on small wire
B = \(\frac{\mu_{0} I}{2 \pi h}\)
The magnetic force on small conductor is , F = BIl sin θ = BIl
Force applied on PQ balance the weight of small current carrying wire.

Long answer type questions

Question 1.
Derive an expression for the force per unit length between two long straight parallel current carrying conductors. Hence define SI unit of current (ampere).
Answer:
Suppose two long thin straight conductors (or wires) PQ and RS are placed parallel to each other in vacuum (or air) carrying currents I₁ and Isub>2 respectively. It has been observed experimentally that when the currents in the wire are in the same direction, they experience an attractive force (fig. a) and when they carry currents in opposite directions, they experience a repulsive force (fig. b).

Let the conductors PQ and RS carry currents I₁ and I₂ in same direction and placed at separation r.

Consider a current-element ‘ab’ of length ΔL of wire RS. The magnetic field produced by current- carrying conductor PQ at the location of other wire RS.
B₁ = \(\frac{\mu_{0} I_{1}}{2 \pi r}\) ………….(1)

According to Maxwell’s right hand rule or right hand palm rule number 1, the direction of B₁ will be perpendicular to the plane of paper and directed downward. Due to this magnetic field, each element of other wire experiences a force. The direction of current element is perpendicular to the magnetic field; therefore the magnetic force on element ab of length ΔL
ΔF = B₁I₂ΔL sin90° = \(\frac{\mu_{0} I_{1}}{2 \pi r}\) I₂ ΔL
∴ The total force on conductor of length L will be
F = \(\frac{\mu_{0} I_{1} I_{2}}{2 \pi r}\) ΔΣL = \(\frac{\mu_{0} I_{1} I_{2}}{2 \pi r}\)L
∴ Force acting per unit length of conductor
f = \(\frac{F}{L}\) = \(\frac{\mu_{0} I_{1} I_{2}}{2 \pi r}\) N/M …………… (2)

According to Fleming’s left hand rule, the direction of magnetic force will be towards PQ, i.e., the force will be attractive.

On the other hand if the currents I₁ and I₂ in wires are in opposite directions, the force will be repulsive. The magnitude of force in each case remains the same.

Definition of SI Unit of Current (Ampere) : In SI system of fundamental unit of current ‘ampere’ is defined assuming the force between the two current carrying wires as standard.

The force between two parallel current carrying conductors of separation r is
\(\frac{F}{L}\) = \(\frac{\mu_{0} I_{1} I_{2}}{2 \pi r}\) N/M

If I₁ = I₂ = 1A, r = lm, then
f = \(\frac{\mu_{0}}{2 \pi}\) = 2 x 10^-7 N/m
Thus, 1 ampere is the current which when flowing in each of parallel conductors placed at separation 1 m in vacuum exert a force of 2 x 10^-7 on 1 m length of either wire.

Question 2.
Draw the labelled diagram of a moving coil galvanometer. Prove that in a radial magnetic field, the deflection of the coil is directly proportional to the current flowing in the coil.
Or (a) Draw a labelled diagram of a moving coil galvanometer.
Describe briefly its principle and working.
(b) Answer the following:
(i) Why is it necessary to introduce a cylindrical soft iron core inside the coil of a galvanometer?
(ii) Increasing the current sensitivity of a galvanometer may not necessarily increase its voltage sensitivity. Explain, giving reason.
Or Explain, using a labelled diagram, the principle and working of a moving coil galvanometer. What is the function of (i) uniform radial magnetic field, (ii) soft iron core?
Or Define the terms (i) current sensitivity and (ii) voltage sensitivity of a galvanometer. Why does increasing the current sensitivity not necessarily increase voltage sensitivity?
Answer:
(a) Moving Coil Galvanometer: A galvanometer is used to detect current in a circuit.

Construction: It consists of a rectangular coil wound on a non-conducting metallic frame and is suspended by phosphor bronze strip between the pole-pieces (N and S) of a strong permanent magnet. A soft iron core in cylindrical form is placed between the coil.

One end of coil is attached to suspension wire which also serves as one terminal (Tx) of galvanometer. The other end of coil is connected to a loosely coiled strip, which serves as the other terminal (T2). The other end of the suspension is attached to a torsion head which can be rotated to set the coil in zero position. A mirror (M) is fixed on the phosphor bronze strip by means of which the deflection of the coil is measured by the lamp and scale arrangement. The levelling screws are also provided at the base of the instrument.

The pole pieces of the permanent magnet are cylindrical so that the magnetic field is radial at any position of the coil.

Principle and Working : When current (I) is passed in the coil, torque τ acts on the coil, given by
τ = NIABsinθ

where θ is the angle between the normal to plane of coil and the magnetic field of strength B, N is the number of turns in a coil.

When the magnetic field is radial, as in the case of cylindrical pole pieces and soft iron core, then in every position of coil the plane of the coil, is parallel to the magnetic field lines, so that θ = 90° and sin 90 ° = 1. The coil experiences a uniform coupler.
Deflecting torque, τ = NIAB
If C is the torsional rigidity of the wire and θ is the twist of suspension strip, then restoring torque = C0. For equilibrium, deflecting torque = restoring torque
i.e., NIAB = Cθ
θ = \(\frac{N A B}{C}\)I ………… (1)
i.e., θ ∝ I
Deflection of coil is directly proportional to current flowing in the coil and hence we can construct a linear scale.

Importance (or Function) of Uniform Radial Magnetic Field
Torque as current carrying coil in a magnetic field is τ = NIAB sinθ In radial magnetic field sinθ = 1, so torque is τ = NIAB.
This makes the deflection (θ) proportional to current. In other words, the radial magnetic field makes the scale linear.

(b)
(i) The cylindrical, soft iron core makes the (1) field radial and (2) increases the strength of the magnetic field, i.e., the magnitude of the torque.

(ii) Sensitivity of Galvanometer
Current sensitivity: It is defined as the deflection of coil per unit current flowing in it.
Sensitivity,
_I = (\(\frac{\theta}{I}\)) = \(\frac{N A B}{C}\)………… (1)
Voltage sensitivity: It is defined as the deflection of coil per unit potential difference across its ends.
i.e., S_V = \(\frac{\theta}{V}\) = \(\frac{N A B}{R_{g} \cdot C}\) …………. (2)
where R_g is resistance of galvanometer.
Clearly for greater sensitivity number of turns N, area A and magnetic field strength B should be large and torsional rigidity C of suspension should be small.
Dividing eqs. (2) by (1)
\(\frac{S_{V}}{S_{I}}=\frac{1}{G}\) = 1 ⇒ S_V = \(\frac{1}{G}\) SI
Clearly, the voltage sensitivity depends on current sensitivity and the resistance of galvanometer. If we increase current sensitivity and resistance G is larger, then it is not certain that voltage sensitivity will be increased. Thus, the increase of current sensitivity does not imply the increase of voltage sensitivity.

Punjab State Board PSEB 12th Class Physics Important Questions Chapter 12 Atoms Important Questions and Answers.

PSEB 12th Class Physics Important Questions Chapter 12 Atoms

Very short answer type questions

Question 1.
Why is the classical (Rutherford) model for an atom of electron orbiting around the nucleus not able to explain the atomic structure?
Answer:
The classical method could not explain the atomic structure as the electron revolving around the nucleus are accelerated and emit energy as the result, the radius of the circular paths goes on decreasing. Ultimately electrons fall into the nucleus, which is not in practice.

Question 2.
Consider two different hydrogen atoms. The electron in each atom is in an excited state. Is it possible for the electrons to have different energies but the same orbital angular momentum according to the Bohr model? (NCERT Exemplar)
Answer:
According to Bohr model electrons having different energies belong to different levels having different values of n. So, their angular momenta will be different, as.
L = \(\frac{n h}{2 \pi} \text { or } L \propto n\)

Question 3.
State Bohr’s quantization condition for defining stationary orbits.
Answer:
According to Bohr’s quantization condition, electrons are permitted to revolve in only those orbits in which the angular momentum of electron is an integral multiple of \(\frac{h}{2 \pi}\) i.e.,
mvr = \(\frac{n h}{2 \pi}\) ,Where n = 1,2,3, ………………
m, y, rare mass, speed, and radius of electron respectively and h being Planck’s constant.

Question 4.
Define ionization energy. What Is Its value for a hydrogen atom?
Answer:
Ionisation Energy: The minimum amount of energy required to remove an electron from the ground state of the atom is known as ionization energy.
Ionisation energy for hydrogen atom =E_∞ – E₁ = – (-13.6 eV) = 13.6 eV

Question 5.
When an electron falls from a higher energy to a lower energy level, the difference in the energies appears in the form of electromagnetic radiation. Why cannot it be emitted as other forms of energy? (NCERT Exemplar)
Answer:
The transition of an electron from a higher energy to a lower energy level can appear in the form of electromagnetic radiation because electrons interact only electromagnetically.

Question 6.
Where is H a -line of the Balmer series in the emission spectrum of hydrogen atom obtained?
Answer:
Hα -line of the Balmer series in the emission spectrum of hydrogen atom is obtained in visible region.

Question 7.
Imagine removing one electron from He⁴ and He³. Their energy levels, as worked out on the basis of Bohr model will be very close. Explain why. (NCERT Exemplar)
Answer:
This is because both the nuclei are very heavy as compared to electron mass.

Question 8.
The mass of H-atom is less than the sum of the masses of a proton and electron. Why is this so? (NCERT Exemplar)
Answer:
Einstein’s mass-energy equivalence gives E – mc².
Thus the mass of an H-atom is m_p + m_e – \(\frac{B}{C^{2}}\)
where B ≈ 13.6 eV is the binding energy. It is less than the sum of masses of a proton and an electron.

Question 9.
Assume that there is no repulsive force between the electrons in an atom but the force between positive and negative charges is given by Coulomb’s law as usual. Under such circumstances, calculate the ground state energy of a He-atom. (NCERT Exemplar)
Answer:
For a He-nucleus with charge 2 e and electrons of charge -e, the energy level in ground state is -E_n = Z²\(\frac{-13.6 \mathrm{eV}}{n^{2}}=2^{2} \frac{-13.6 \mathrm{eV}}{1^{2}}\)= -54.4eV
Thus, the ground state will have two electrons each of energy E and the total ground state energy would be -(4 x 13.6) eV = -54.4 eV.

Question 10.
Would the Bohr formula for the H-atom remain unchanged if proton had a charge (+ 4/3)e and electron a charge (-3/4)e, where e = 1.6 x 10^-19 C ? Give reasons for your answer. (NCERT Exemplar)
Answer:
Yes, since the Bohr formula involves only the product of the charges.

Short answer type questions

Question 1.
In an experiment of α-particle scattering by a thin foil of gold, draw a plot showing the number of particles scattered versus the scattering angle θ. Why is it that a very small fraction of the particles are scattered at θ > 90°?

Answer:
A small fraction of the alpha particles scattered at angle θ > 90° is due to the reason. That if impact parameter ‘b’ reduces to zero, coulomb force increases, hence alpha particles are scattered at angle θ>9O°, and only one alpha particle is scattered at angle 180°.

Question 2.
(i) State Bohr postulate of hydrogen atom that gives the relationship for the frequency of emitted photon in a transition,
(ii) An electron jumps from fourth to first orbit in an atom. How many maximum number of spectral lines can be emitted by the atom? To which series these lines correspond?
Answer:
(i) Bohr’s Third Postulate: It states that an electron might make a transition from one of its specified non-radiating orbits to another of lower energy. When it does so, a photon is emitted having energy equal to the energy difference between the initial and final states. The frequency of the emitted photon is given by
hv = E_i – E_f
Where E_i and E_f are the energies of the initial and final states and E_i > E_f.
(ii) Electron jumps from fourth to first orbit in an atom

∴ Maximum number of spectral lines can be
4c₂ = \(\frac{4 !}{2 ! 2 !}=\frac{4 \times 3}{2}\) = 6
The line responds to Lyman series (e^– jumps to 1st orbit), Balmer series (e^– jumps to 2nd orbit), Paschen series (e^– jumps to 3rd orbit).

Question 3.
Using de Broglie’s hypothesis, explain with the help of a suitable diagram, Bohr’s second postulate of quantization of energy levels in a hydrogen atom.
Answer:
According to de Broglie’s hypothesis.
λ = \(\frac{h}{m v}\) ……………………….. (i)
According to de Broglie’s condition of stationary orbits, the stationary orbits are those which contain complete de Broglie wavelength.
2πr = nλ ………………………….. (2)

Substituting value of λ from eq. (2) in eq. (1), we get
2πr = n \(\frac{h}{m v}\)
⇒ mvr = n \(\frac{h}{2 \pi}\) ………………………… (3)
This is Bohr’s postulate of quantisation of energy levels.

Question 4.
In the study of Geiger-Marsden experiment on scattering of α-particles by a thin foil of gold, draw the trajectory of a-particles in the coulomb field of target nucleus. Explain briefly how one gets the information on the size of the nucleus from this study. From the relation R = R₀ A^1/3, where, R₀ is constant and A is the mass number of the nucleus, show that nuclear matter density is independent of A.
Answer:

From this experiment, the following is observed :
1. Most of the α-particles pass straight through the gold foil. It means that they do not suffer any collision with gold atoms.
2. About one α-particle in every 8000 α-particles deflects by more than 90°. As most of the a-particles gounder flected and only a- few get deflected, this shows that most of the space in an atom is empty and at the center of the atom, there exists a nucleus.

By the number of a-particles get deflected, the information regarding size of the nucleus can be known.
If m is the average mass of the nucleus and R is the nuclear radius, then mass of nucleus = mA, where A is the mass number of the element. Volume of the nucleus,

This shows that the nuclear density is independent of A.

Question 5.
Show that the first few frequencies of light that is emitted when electrons fall to nth level from levels higher than n, are approximate harmonics (L e., in the ratio 1: 2: 3,…) when n>> 1. (NCERTExempiar)
Answer:
The frequency of any line in a series in the spectrum of hydrogen-like atoms corresponding to the transition of electrons from (n + p) level to nth level can be expressed as a difference of two terms:
V_mn = \(c R Z^{2}\left[\frac{1}{(n+p)^{2}}-\frac{1}{n^{2}}\right] \)
where, m=n+p,(p=1,2,3,…………………………..)
and R is Rydberg constant.
For p << n

V_mn = \(c R Z^{2}\left[\frac{1}{n^{2}}\left(1+\frac{p}{n}\right)^{-2}-\frac{1}{n^{2}}\right]\)

Thus, the first few frequencies of light that is emitted when electrons fall to the nth level from levels higher than n, are approximate harmonic (i. e., in the ratio 1:2:3,…) when n>>1.

Long answer type questions

Question 1.
Using the postulates of Bohr’s model of hydrogen atom, obtain an expression for the frequency of radiation emitted when atom make a transition from the higher energy state with quantum number n₁ to the lower energy state with quantum number n_f (n_f < n_i).
Or
Using Bohr’s postulates, obtain the expression for the total energy of the electron in the stationary states of the hydrogen atom. Hence draw the energy level diagram showing how the line spectra corresponding to Balmer series occur due to transition between energy levels.
Or
Using Rutherford model of the atom, derive the expression for the total energy of the electron in hydrogen atom. What is the significance of total negative energy possessed by the electron?
Answer:
Suppose m be the mass of an electron and v be its speed in nth orbit of radius r. The centripetal force for revolution is produced by electrostatic attraction between electron and nucleus.
\(\frac{m v^{2}}{r}=\frac{1}{4 \pi \varepsilon_{0}} \frac{(Z e)(e)}{r^{2}}\) …………………… (1) [Form rutherford Model]
or mv² = \(\frac{1}{4 \pi \varepsilon_{0}} \frac{Z e^{2}}{r}\)
So, Kinetic energy E_k = \(\frac{1}{2} m v^{2}\)
E_k = \(\frac{1}{4 \pi \varepsilon_{0}} \frac{Z e^{2}}{2 r}\)
Potential energy (PE) = \(\frac{1}{4 \pi \varepsilon_{0}} \frac{(Z e)(-e)}{r}=-\frac{1}{4 \pi \varepsilon_{0}} \frac{Z e^{2}}{r}\)
Total energy E = \(E_{K}+P E=\frac{1}{4 \pi \varepsilon_{0}} \frac{Z e^{2}}{2 r}+\left(-\frac{1}{4 \pi \varepsilon_{0}} \frac{Z e^{2}}{r}\right)\)
= \(-\frac{1}{4 \pi \varepsilon_{0}} \frac{Z e^{2}}{2 r}\)

For nth orbit, E can be written as E_n
so,E_n = \(-\frac{1}{4 \pi \varepsilon_{0}} \frac{Z e^{2}}{2 r_{n}}\) …………………. (2)
Negative sign indicates that the electron remains bound with the nucleus (or electron-nucleus form an attractive system) From Bohr’s postulate for quantisation of angular momentum.
mvr = \(\frac{n h}{2 \pi}\)
⇒ v = \(\frac{n h}{2 \pi m r} \)
Substituting this value of v in equation (1), we get

For Bohr’s radius, n = 1
Substituting value of r_n in equation (2), we get
E_n = \(\frac{1}{4 \pi \varepsilon_{0}} \frac{Z e^{2}}{2\left(\frac{\varepsilon_{0} h^{2} n^{2}}{\pi m Z e^{2}}\right)}=-\frac{m Z^{2} e^{4}}{8 \varepsilon_{0} h^{2} n^{2}}\)
R is called Rydberg constant.

For hydrogen atom Z =1, E_n = \(\frac{-R c h}{n^{2}}\)
If n_i and n_f are the quantum numbers of initial and final states and E_i and
E_f are energies of electrons in H-atoms in initial and final state, we have

For Balmer series, n_f=2, while n_i =3, 4, 5, …… ∞.

Punjab State Board PSEB 12th Class Physics Important Questions Chapter 3 Current Electricity Important Questions and Answers.

PSEB 12th Class Physics Important Questions Chapter 3 Current Electricity

Very short answer type questions

Question 1.
Two identical cells, each of emf E, having negligible internal resistance, are connected in parallel with each other across an external resistance R. What is the current through this resistance?
Answer:

⇒ I = \(\frac{E_{\text {eq. }}}{R+r_{\text {eq. }}}\)
Given,internal resistance, r = 0
∴ I = \(\frac{E_{\mathrm{eq} .}}{R}\)

Question 2.
Define mobility of a charge carrier. What is its relation with relaxation time?
Answer:
It is defined as how fast electron moves from one place to another.
It is also defined as drift velocity per unit electric field. The SI unit of mobility is m²/V-sec and it is denoted as μ.
μ = \(\frac{\left|v_{d}\right|}{E}=\frac{e E \tau}{m E}=\frac{e \tau}{m}\)
⇒ μ ∝ τ

Question 3.
Define the term ‘relaxation time’ in a conductor.
Answer:
The average time between successive collisions of electrons conductor is known as relaxation time.

Question 4.
Write the expression for the drift velocity of charge carriers in a conductor of length ‘L’ across which a potential difference ‘V’ is applied.
Answer:
v_d = \(\frac{e V}{m L}\)τ

Question 5.
For household electrical wiring, one uses Cu wires or Al wires. What considerations are kept in mind? (NCERT Exemplar)
Answer:
Two considerations are required: (i) cost of metal, and (ii) good conductivity of metal. Cost factor inhibits silver. Cu and Al are the next best conductors.

Question 6.
Define the current sensitivity of a galvanometer. Write its SI unit.
Answer:
Ratio of deflection produced in the galvanometer and the current flowing through it is called current sensitivity.
Current sensitivity S_i = \(\frac{\theta}{I}\)
SI unit of current sensitivity S_i is division/ampere or radian/ampere.

Question 7.
Nichrome and Copper wires of same length and same radius are connected in series. Current I is passed through them. Which wire gets heated up more? Justify your answer.
Answer:
Nichrome, since its resistance is high.

Question 8.
Why are alloys used for making standard resistance coils?
(NCERT Exemplar)
Answer:
Alloys have:

• low value of temperature coefficient and the resistance of the alloy does not vary much with rise in temperature.
• high resistivity, so even a smaller length of the material is sufficient to design high standard resistance.

Question 9.
What is the advantage of using thick metallic strips to join wires in a potentiometer? (NCERT Exemplar)
Answer:
The metal strips have low resistance and need not be counted in the potentiometer length l of the null point. One measures only their lengths along the straight segments (of length 1 metre each). This is easily done with the help of centimeter rulings or meter ruler and leads to accurate measurements.

Question 10.
Is the motion of a charge across junction momentum conserving? Why or why not? (NCERT Exemplar)
Answer:
When an electron approaches a junction, in addition to the uniform electric field E facing it normally, it keep the drift velocity fixed as drift velocity depend on E by the relation
V_d = \(\frac{e E \tau}{m}\)
This result into accumulation of charges on the surface of wires at the junction. These produce additional field. These fields change the direction of momentum.
Thus, the motion of a charge across junction is not momentum conserving.

Short answer type questions

Question 1.
Sketch a graph showing the variation of resistivity of carbon with temperature.
Or Plot a graph showing temperature dependence of resistivity for a typical semiconductor. How is this behaviour explained?
Answer:
The resistivity of a typical semiconductor (carbon) decreases with increase of temperature. The graph is shown in figure.

Explanation: In semiconductor the number density of free electrons (n) increases with increase in temperature (T) and consequently the relaxation period decreases. But the effect of increase in n has higher impact than decrease of τ. So, resistivity decreases with increase in temperature.

Question 2.
Two cells of emf ε₁ and ε₂ having internal resistances r₁ and r₂ respectively are connected in parallel as shown. Deduce the expressions for the equivalent emf and equivalent internal resistance of a cell which can replace the combination between the points B₁ and B₂.

Answer:
Consider a parallel combination of the cells. I₁ and I₂ are the currents leaving the positive electrodes of the cells. At point B₁, I₁ and I₂ flow in whereas current I flows out. Therefore, we have
I = I₁ + I₂ …………….. (1)
Let V(B₁) and V(B₂) be the potentials at B₁ and B₂ respectively.
Then, considering the first cell, the potential difference across its terminals is V(B₁) – V(B₂). Hence, from equation V = E – Ir we have
V = V(B₁) – V(B₂) = E₁ – I₁r₁ …………… (2)
Points B₁ and B₂ are connected exactly Similarly to the second cell. Hence, considering the second cell, we also have
V = V(B₁) – V(B₂)
= E₂ – I₂r₂ …………… (3)
Combining equations (1), (2) and (3), we have

If we want to replace the combination by a single cell, between Bl and B₂, of emf E_eq and internal resistance r_eq, we would have
V = E_eq – Ir_eq

Question 3.
State Kirchhoff s rules of current distribution in an electrical network.
Or State KirchhofPs rules. Explain briefly how these rules are justified.
Answer:
Junction Rule: In an electric circuit, the algebraic sum of currents at any junction is zero.
At any junction, the sum of the currents entering the junction is equal to the sum of currents leaving the junction.
ΣI = 0
Justification: This rule is based on the law of conservation of charge.
Loop Rule: The algebraic sum of changes in potential around any closed loop involving resistors and cells in the loop must be zero.
ΣΔV = 0
or The algebraic sum of emf s in any loop of a circuit is equal to the
sum of products of currents and resistances in it.
ΣΔE = ΣIR
Justification: This rule is based on the law of conservation of energy,

Question 4.
Define the term current density of a metallic conductor. Deduce the relation connecting current density (J) and the conductivity σ of the conductor, when an electric field E, is applied to it.
Answer:
Current density at a point in a conductor is defined as the amount of current flowing per unit area of the conductor around that point provided the area is held in a direction normal to the current,
J = \(\frac{I}{A}\)
Current density is a vector quantity. Its direction is the direction of motion of positive charge. The unit of current density is ampere/metre² or [Am^-2].
Relation between J, σ and E

Question 5.
What is Wheatstone bridge? Deduce the condition for which Wheatstone bridge is balanced.
Or The given figure shows a network of resistances R₁, R₂, R₃ and R₄.

Using Kirchhoffs laws, establish the balance condition for the network.
Or Use Kirchhoffs law to obtain the balance Wheatstone’s bridge.
Answer:

The Wheatstone bridge is an arrangement of four resistances. In this bridge, four resistances are connected on four arms of quadrilateral. In one diagonal, a battery and a key are connected. In second diagonal a galvanometer is connected as shown in fig. Consider P,Q,R and S four resistances are connected on the sidesAB,BC, AD and DC of the quadrilateral respectively.

Galvanometer G is connected between points B and D and a battery E is connected between A and C. Now in balance condition, when the deflection in a galvanometer is zero in closed mesh ABDA, then by applying Kirchhoffs law,
I₁p – IR = 0
or I₁P = I₂R ………….. (1)
In closed mesh CBDC,
I₁Q = I₂S ……………… (2)
Dividing (1) by (2) \(\frac{P}{Q}=\frac{R}{S}[latex]
This is the balanced condition of the Wheatstone bridge.

Question 6.
First a set of n equal resistors of R each are connected in series to a battery of emf E and internal resistance R. A current I is observed to flow. Then the n resistors are connected in parallel to the same battery. It is observed that the current is increased 10 times. What is n ? (NCERT Exemplar)
Answer:
When n resistors are in series, I = [latex]\frac{E}{R+n R}\) ;
When n resistors are in parallel, \(\frac{E}{R+\frac{R}{n}}\) 10I
\(\frac{1+n}{1+\frac{1}{n}}\) = 10 ⇒ \(\frac{1+n}{n+1}\) n = 10
∴ n = 10

Question 7.
Two cells of same emf E but internal resistance r and r₁ and r₂ are connected in series to an external resistor R (Fig.). What should be the value of R so that the potential difference across the terminals of the first cell becomes zero. NCERT Exemplar)
Answer:
I = \(\frac{E+E}{R+r_{1}+r_{2}}\)
V₁ = E – Ir₁ = E – \(\frac{2 E}{r_{1}+r_{2}+R}\) r₁ = 0

Long answer type questions

Question 1.
(i) Find the magnitude and direction of current in 1Ω resistor in given circuit.

(ii)Two students X and Y perform an experiment on potentiometer separately using the circuit diagram shown below.
Keeping other things unchanged (a) X increases the value of resistance R, (b) Y decreases the value of resistance S in the set up. How will these changes affect the position of null point in each case and why?

Answer:
(i) For the mesh APQBA
-6 -1 (I₂ – I₁) + 3I₁ = 0
or -I₂ + 4I₁ = 6 …………… (1)

For the mesh PCDQP
2I₂ – 9 + 3I₂ + 1(I₂ – I₁) = 0
or 6I₂ – I₁ = 9 …………… (2)
Solving eqs. (1) and (2), we get
I = \(\frac{45}{23}\) A
I= \(\frac{42}{23}\) A
∴ Current through the 1Ω resistor = (I₂ – I₁) = \(\) A

(ii) (a) By increasing resistance R, the current in main circuit decreases, so potential gradient decreases. Hence, a greater length of wire would be needed for balancing the same potential difference. So, the null point would shift towards right (i.e., towards B).

(b) By decreasing resistance S, the terminal potential difference V = \(\frac{E}{1+\frac{r}{S}}\) across cell decreases, so balance is obtained at small length i.e., point will be obtained at smaller length. So, the null point would shift towards left (i.e., towards A).

Question 2.
A room has AC run for 5 hours a day at a voltage of 220 V. The wiring of the room consists of Cu of 1 mm radius and a length of 10m. Power consumption per day is 10 commercial units. What fraction of it goes in the joule heating in wires? What would happen if the wiring is made of aluminium of the same dimensions? [ρCu = 1.7 × 10^-8Ωm, ρAL = 2.7 × 10^-8Ωm] (NCERT Exemplar)
Answer:
Power consumption in a day i.e., in 5 hours = 10 units
Or power consumption per hour = 2 units
Or power consumption = 2 units = 2 kW = 2000 W
Also, we know that power consumption in resistor,
P = V × I
⇒ 2000 W = 220 V × I
or I = 9 A
Now, the resistance of wire is given by R = ρ \(\frac{l}{A}\)
where, A is cross-sectional area of conductor. Power consumption in first current carrying wire is given by
P = I² . R
ρ \(\frac{l}{A}\) I² = 1.7 × 10^-8 × \(\frac{10}{\pi \times 10^{-6}}\) × 81 = 4J/s A
The fractional loss due to the joule heating in first wire
= \(\frac{4}{2000}\) × 100 = 0.2%
Power loss in Al wire = 4\(\frac{\rho_{A l}}{\rho_{C u}}\) = 1.6 × 4 = 6.4 J/s
The fractional loss due to the joule heating in second wire
= \(\frac{6.4}{2000}\) × 100 =0.32%

Punjab State Board PSEB 12th Class Physics Important Questions Chapter 13 Nuclei Important Questions and Answers.

PSEB 12th Class Physics Important Questions Chapter 13 Nuclei

Very short answer type questions

Question 1.
Write the relationship between the size of a nucleus and Its mass number (A).
Answer:
The relationship is R = RoA^1/3
where, R = radius of nucleus, A = mass number.

Question 2.
Define the activity of a given radioactive substance. Write its SI unit.
Answer:
The activity of a sample is defined as the rate of disintegration taking place in the sample of radioactive substances.
SI unit of activity is Becquerel (Bq).
1 Bq = 1 disintegration/second

Question 3.
Why is it found experimentally difficult to detect neutrinos in nuclear p-decay?
Answer:
Neutrinos are difficult to detect because they are massless, have no charge, and do not interact with nucleons.

Question 4.
A nucleus undergoes p-decay. How does its :
(i) mass number
(ii) atomic number change?
Answer:
During p-decay,
(i) no change in mass number.
(ii) atomic number increases by 1.

Question 5.
In pair annihilation, an electron and a positron destroy each other to produce gamma radiations. How is the momentum conserved? (NCERT Exemplar)
Answer:
In pair annihilation, an electron and a positron destroy each other to produce 2γ photons which move in opposite directions to conserve linear momentum.
The annihilation is shown below ₀e⁺¹ + ₀ e⁺¹ →2γ ray photons.

Question 6.
Which one of the following cannot emit radiation and why? Excited nucleus, excited electron. (NCERT Exemplar)
Answer:
Excited electrons cannot emit radiation because energy of electronic energy levels is in the range of eV and not MeV (mega electron volt). y-radiations have energy of the order of MeV.

Question 7.
He_2³ and He_1³ nuclei have the same mass number.
Do they have the same binding energy? (NCERT Exemplar)
Answer:
Nuclei He_2³ and He_1³ have the same mass number.
He_2³ has two protons and one neutron. He_1³ has one proton arid two neutrons.
The repulsive force between protons is missing in ₁He³, so the binding energy of ₁He³ is greater than that of ₂He³.

Question 8.
Which sample, A or B as shown in figure has shorter mean-life? (NCERT Exemplar)

Answer:
B has shorter mean life as λ is greater for B.

Short answer type questions

Question 1.
(i) Define the terms (a) half-life (b) average life. Find out the relationship with the decay constant (λ).
(ii) A radioactive nucleus has a decay constant ) λ = 0.346 (day)^-1
How long would it take the nucleus of decay to 75% of Its Initial amount?
Answer:
(i)
(a) Half-life of a radioactive element is defined as the time during which half number of atoms present initially in the sample of the element decay or it is the time during which number of atoms left undecayed in the sample is half the total number of atoms present in the sample.
it is represented by T_1/2.
From the equation N = N₀e^-λt ,
At half-life, t = T_1/2,N = \(\frac{N_{0}}{2}\)

On taking log both sides, we get
λT_1/2 = log _e2
T _1/2 = \(\frac{\log _{e} 2}{\lambda}=\frac{\log _{10} 2 \times 2.303}{\lambda}\)
= \(\frac{0.3010 \times 2.303}{\lambda}\)
After n half-life, the number of atoms left undecayed is given by
N = N₀\(\left(\frac{1}{2}\right)^{n}\)
T_1/2 = \(\frac{0.6932}{\lambda}\)

(b) Average life of a radioactive element can be obtained by calculating the total life time of all atoms of the element and dividing it by the total number of atoms present initially in the sample of the element.
Average life or mean life of radioactive element is
τ = \(\frac{\text { Total life of all atoms }}{\text { Total number of atoms }}\)
τ = \(=\int_{0}^{N_{0}} \frac{t d N}{N_{0}}=\int_{\infty}^{0-\lambda N_{0} e^{-\lambda t} d t \times t}{N_{0}}\)
[when N =N₀,t = 0 and when N = 0, t = ∞] [∵dN =-λ(N₀e^–λt)dt]
= λ₀∫^∞ te ^-λtdt

(ii) Given, λ = 0.3465 (day)
According to the radioactive decay law, we have
R = R₀e^-λt

⇒ t = 0.830 s

Question 2.
(i) Write three characteristic properties of nuclear force.
(ii) Draw a plot of potential energy of a pair of nucleons as a function of their separation. Write two important conclusions that can be drawn from the graph.
Answer:
(i) Characteristic Properties of Nuclear Force
(a) Nuclear force act between a pair of neutrons, a pair of protons and also between a neutron-proton pair, with the same strength. This shows that nuclear forces are independent of charge.
(b) The nuclear forces are dependent on spin or angular momentum of nuclei.
(c) Nuclear forces are non-central forces. This shows that the distribution of nucleons in a nucleus is not spherically symmetric. From the plot, it is concluded that

(ii)
(a) The potential energy is minimum at a distance r₀(=0.8fm) which means that the force is attractive for distance larger than 0.8 fm and repulsive for the distance less than 0.8 fm between the nucleons.
(b) Nuclear forces are negligible when the distance between the nucleons is more than 10 fm.

Question 3.
Explain giving necessary reaction, how energy is released during:
(i) fission
(ii) fusion
Answer:
(i) Nuclear Fission: The phenomenon of splitting of heavy nuclei (mass number > 120) into smaller nuclei of nearly equal masses is known as nuclear fission. In nuclear fission, the sum of the masses of the product is less than the sum of masses of the reactants. This difference of mass gets converted into energy E = me and hence sample amount of energy is released in a nuclear fission.
e.g., _92²³⁵ U + ₀¹n → ₅₆¹⁴¹Ba + _36⁹²Kr + 3 _0¹ + Q
Masses of reactant = 235.0439 amu + 1.0087 amu
= 236.0526 amu
Masses of product = 140.9139 + 91.8973 + 3.0261
= 235.8373 amu
Mass defect = 236.0526 -235.8373
= 0.2153 amu
∵ 1 amu = 931MeV
⇒ Energy released = 0.2153 x 931
⇒ 200 MeV nearly

(ii) Nuclear Fusion: The phenomenon of conversion of two lighter nuclei into a single heavy nucleus is called.nuclear fusion. Since the mass of the heavier product nucleus is less than the sum of masses of reactant nuclei and therefore certain mass defect occurs which converts into energy as per Einstein’s mass-energy relation. Thus, energy is released during nuclear fusion.
e.g., ₁H¹ + ₁H¹ → ₁H² + e⁺ + v + 0.42 MeV
Also, ₁H² + ₁H² → ₁H³ + ₁H¹ + 4.03 MeV

Question 4.
Give reasons for :
(a) Lighter elements are better moderators for a nuclear reactor than heavier elements.
(b) In a natural uranium reactor, heavy water is preferred moderator as compared to ordinary water.
(c) Cadmium rods are provided in a reactor.
(d) Very high temperatures as those obtained in the interior of the sun are required for fusion reaction.
Answer:
(a) A moderator slows down fast neutrons released in a nuclear reactor. The basic principle of mechanics is that the energy transfer in a collision is the maximum when the colliding particles have equal masses. As lighter elements have mass close to that of neutrons, lighter elements are better moderators than heavier elements.

(b) Ordinary water has hydrogen nuclei (₁¹H) which have greater absorption capture for neutrons; so ordinary water will absorb neutrons rather than slowing them; on the other hand, the heavy hydrogen nuclei (₂¹H) have negligible absorption capture for neutrons, so they share energy with neutrons and neutrons are slowed down.

(c) Cadmium has high absorption capture for neutrons; so cadmium rods are used to absorb extra neutrons; so nuclear fission in a nuclear reactor is controlled; therefore cadmium rods are called control rods.

(d) In nuclear fusion, two positively charge nuclei fuse together. When two positively charged nuclei come near each other to fuse together, strong electrostatic repulsive force acts between them.
To overcome this repulsive force, extremely high temperatures of the order of 10⁸ K are required.
This may be calculated as follows: For fusion of H-nuclei,

The temperature in the interior of sun is about 2 x 10⁷ K.
Therefore, very high temperatures of the order 10⁷ K are required for fusion reaction to take place.

Question 5.
Deuteron is a bound state of a neutron and a proton with a binding energy B 2.2 Mev. A γ-ray of energy E is aimed at a deuteron nucleus to try to break it into a (neutron + proton) such that the n and p move in the direction of the incident y-ray. If E = B, show that this cannot happen. Hence, calculate how much bigger than B must be E be for such a process to happen. (NCERT Exemplar)
Answer:
Given binding energy B = 2.2 MeV
From the energy conservation law,
E-B=K_n+K_p = \(\frac{p_{n}^{2}}{2 m}+\frac{p_{p}^{2}}{2 m}\) ………………….. (1)
From conservation of momentum,
P_n +P_p = \(\frac{E}{c}\)
As E = B,Eq. (1) p_n²+ P_p² =0 ……………………………. (2)
It only happen if p_n = p_p = 0

So, the Eq. (2) cannot satisfied and the process cannot take place.
Let E = B +X, when X<< B for the process to take place.
Put value of p1 from Eq. (2) in Eq. (1), we get
X = \(\frac{\left(\frac{E}{c}-p_{p}\right)}{2 m}+\frac{p_{p}^{2}}{2 m}\)
or 2p_p² – \(\frac{2 E p_{p}}{c}+\frac{E^{2}}{c^{2}}\) – 2mx = 0

Using the formula of quadratic equation, we get
P_p = \(\frac{\frac{2 E}{c} \pm \sqrt{\frac{4 E^{2}}{c^{2}}-8\left(\frac{E^{2}}{c^{2}}-2 m X\right)}}{4}\)
For the real value p_p, the discriminant is positive

Long answer type questions

Question 1.
Define the term: Half-life period and decay constant of a radioactive sample. Derive the relation between these terms.
Answer:
Half-life Period: The half-life period of an element is defined as the time in which the number of radioactive nuclei decay to half of its initial value.
Decay Constant: The decay constant of a radioactive element is defined as the reciprocal of time in which the number of undecayed nuclei of that radioactive element falls to times of its initial value. Relation between Half-life and Decay Constant:
The radioactive decay equation is N = N₀e^-λt …………………………. (i)
When t = T,N= \(\frac{N_{0}}{2}\)
∴ \(\frac{N_{0}}{2}\) = N₀e^-λT or e^-λT = \(\frac{1}{2}\)
……………………. (2)
Taking log on both sides, we get

Question 2.
Draw the graph showing the variation of binding energy per nucleon with the mass number for a large number of nuclei 2 < A < 240. What are the main inferences from the graph? How do you explain the constancy of binding energy in the range 30<A<170 using the property that the nuclear force is short-ranged? Explain with the help of this plot the release of energy in the processes of nuclear fission and fusion.
Answer:
The variation of binding energy per nucleon versus mass number is shown in figure.

Inferences from Graph
1. The nuclei having mass numbers below 20 and above 180 have relatively small binding energy and hence they are unstable.
2. The nuclei having mass numbers 56 and about 56 have maximum binding energy – 5.8 MeV and so they are most stable.
3. Some nuclei have peaks, e.g., ₂He⁴, ₆C¹², ₈O¹⁶; this indicates that these nuclei are relatively more stable than their neighbors.
(i) Explanation of constancy of binding energy: Nuclear force is short-ranged, so every nucleus interacts with its neighbors only, therefore binding energy per nucleon remains constant.

(ii) Explanation of nuclear fission: When a heavy nucleus (A ≥ 235 say) breaks into two lighter nuclei (nuclear fission), the binding energy per nucleon increases i.e., nucleons get more tightly bound. This implies that energy would be released in nuclear fission,

(iii) Explanation of nuclear fusion: When two very light nuclei (A ≤ 10) join to form a heavy nucleus, the binding energy per nucleon of fused heavier nucleus is more than the binding energy per nucleon of lighter nuclei, so again energy would be released in nuclear fusion.

Punjab State Board PSEB 12th Class Physics Important Questions Chapter 2 Electrostatic Potential and Capacitance Important Questions and Answers.

PSEB 12th Class Physics Important Questions Chapter 2 Electrostatic Potential and Capacitance

Very short answer type questions

Question 1.
A charge ‘q’ is moved from a point A above a dipole of dipole moment ‘p’ to a point B below the dipole in equatorial plane without acceleration. Find the work done in the process.

Answer:
Work done in the process is zero. Because, equatorial plane of a dipole is equipotential surface and work done in moving charge oh equipotential surface is zero.
W = qV_AB = q × 0 = 0

Question 2.
A point charge Q is placed at point O as shown in the figure. Is the potential difference (V_A – V_B) positive, negative or zero if Q is
(i) positive
(ii) negative

Answer:
Let the distance of points A and B from charge Q be r_A and r_B, respectively.
∴ Potential difference between points A and B
V_A – V_B = \(\frac{q}{4 \pi \varepsilon_{0}}\left[\frac{1}{r_{A}}-\frac{1}{r_{B}}\right]\)
As r_A = OA, r_B = r_B = OB
and r_A < r_B, \(\frac{1}{r_{A}}>\frac{1}{r_{B}}\)
There,[latex]\frac{1}{r_{A}}-\frac{1}{r_{B}}[/latex] has positive value.
(V_A – V_B) depends on the nature of charge q.
(V_A – V_B) is positive when Q > 0
(V_A – V_B ) is negative when Q < 0

Question 3.
A hollow metal sphere of radius 5 cm is charged such that potential on its surface is 10 V. What is the potential at the centre of the sphere?
Answer:
The electric potential at every point inside the charged spherical shell is same and equal to the electric potential on its surface.
The electric potential at the centre of sphere is 10 V.

Question 4.
Can two equipotential surfaces intersect each other? Justify your answer.
No, two equipotential surfaces cannot intersect each other because

1. Two normals can be drawn at intersecting point on two surfaces which gives two directions of E at the same point which is impossible.
2. Also two values of potential at the same point is not possible.

Question 5.
Why electrostatic potential is constant throughout the volume of the conductor and has the same value as on its surface?
Since, electric field intensity inside the conductor is zero. So, electrostatic potential is a constant.
But, E = –\(\frac{\Delta V}{\Delta r}\)
∴ E = 0, ΔV = 0
or V₂ – V₁ =0, V₂ – V₁
The potential at every point inside the conductor remains same.

Question 6.
In a certain 0.5 cm³ of space, electric potential is found to be 7 V throughout. What is the electric field in this region?
Answer:
Zero, because electric potential is same throughout as
E = \(\frac{d V}{d r}\)

Question 7.
Distinguish between a dielectric and a conductor.
Answer:
Dielectrics are non-conductors and do not have free electrons at all. While conductor has free electrons in its any volume which makes it able to pass the electricity through it.

Question 8.
The given graph shows the variation of charge q versus potential difference V for two capacitors C₁ and C₂. Both the capacitors have same plate separation but plate area of C₂ is greater than that C₁. Which line (A or B) corresponds to C₁ and why?

Answer:
Line B corresponds to Q because slope (q/V) of B is less than slope of A.

Question 9.
Do free electrons travel to region of higher potential or lower potential? (NCERT Exemplar)
Answer:
Free electrons would travel to regions of higher potentials as they are negatively charged.

Question 10.
Can there be a potential difference between two adjacent conductors carrying the same charge? (NCERT Exemplar)
Answer:
Yes, if the sizes are different.

Question 11.
Prove that, if an insulated, uncharged, conductor is placed near a charged conductor and no other conductors are present, the uncharged body must intermediate in potential between that of the charged body and that of infinity. (NCERT Exemplar)
Answer:
Let us take any path from the charged conductor to the uncharged conductor along the direction of electric field. Therefore, the electric potential decrease along this path.
Now, another path from the uncharged conductor to infinity will again continually lower the potential further. This ensures that the uncharged body must be intermediate in potential between that of the charged body and that of infinity.

Question 12.
A test charge q is made to move in the electric field of a point charge Q along two different closed paths [figure]. First path has sections along and perpendicular to lines of electric field.
Second path is a rectangular loop of the same area as the first loop. How does the work done compare in the two cases? (NCERT Exemplar)

Answer:
As electric field is conservative, work done will be zero in both the cases.

Short answer type questions

Question 1.
What is electrostatic shielding? How is this property used in actual practice? Is the potential in the cavity of a charged conductor zero?
Answer:
Whatever be the charge and field configuration outside, any cavity in a conductor remains shielded from outside electric influence. The field inside a conductor is zero. This is known as electrostatic shielding.

• Sensitive instruments are shielded from outside electrical influences by enclosing them in a hollow conductor.
• During lightning it is safest to sit inside a car, rather than near a tree. The metallic body of a car becomes an electrostatic shielding from lightening.

Potential inside the cavity is not zero. Potential is constant.

Question 2.
Two capacitors of unknown capacitances C₁ and C₂ are connected first in series and then in parallel across a battery of 100 V. If the energy stored in the two combinations is 0.045 J and 0.25 J respectively, determine the value of C₁ and C₂. Also calculate the charge on each capacitor in parallel combination.
Answer:
Energy stored in a capacitor,

Q₁ = C₁V = 38.2 × 10^-6 × 100 = 38.2 × 10^-4 C
Q₂ = C₂V = 11.8 × 10^-6 × 100 = 11.8 × 10^-4C

Question 3.
Calculate the equivalent capacitance between points A and B in the circuit below. If a battery of 10 V is connected across A and B, calculate the charge drawn from the battery by the circuit.

Answer:

Question 4.
A parallel plate capacitor is charged by a battery to a potential. The battery is disconnected and a dielectric slab is inserted to completely fill the space between the plates.
How will
(a) its capacitance
(b) electric field between the plates and
(c) energy stored in the capacitor be affected? Justify your answer giving necessary mathematical expression for each case.
Answer:
On introduction of dielectric slab in a isolated charged capacitor.
(a) The capacitance (C’) becomes K times of original capacitor as
C = \(\frac{\varepsilon_{0} A}{d}\)
and C ‘ = \(\frac{K \varepsilon_{0} A}{d}\)

(b) The total charge on the capacitor remains conserved on introduction of dielectric slab. Also, the capacitance of capacitor increases to K times of original values.
∴ CV = C’V’
CV = (KC)V’
⇒ V’ = \(\frac{V}{K}\)
∴ New electrical field,
E’ = \(\frac{V^{\prime}}{d}=\left(\frac{V / K}{d}\right)=\left(\frac{V}{d}\right) \frac{1}{K}=\frac{E}{K}\)
∴ On introduction of dielectric medium new electric field E’ becomes \(\) times of its original value.
(c) Energy stored initially,
U = \(\frac{q^{2}}{2 C}\)
Energy stored later,
U’ = \(\frac{q^{2}}{2(K C)}\) [< C’ = KG]
where, K = dielectric constant of medium
⇒ U’ = \(\frac{1}{K}\left(\frac{q^{2}}{2 C}\right)\)
⇒ U’ = \(\frac{1}{K}\) × U
The energy stored in the capacitor decreases and becomes \(\frac{1}{K}\) times of original energy.

Question 5.
Two identical parallel plate capacitors A and B are connected to a battery of V volts with the switch S closed. The switch is now opened and the free space between the plates of the capacitors is filled with a dielectric of dielectric constant K. Find the ratio of the total electrostatic energy stored in both capacitors before and after the introduction of the dielectric.

Answer:
Let C be the capacitance of each capacitor.
With switch S closed, the two capacitor are in parallel.
∴ Equivalent capacitance is 2 C.
∴ Energy stored = \(\frac{1}{2}\)(2C)V²
U₁ = CV² …………… (1)

Now, when switch is opened and then free space of capacitors are filled with dielectric, the capacitance of each capacitor will be KC. For capacitor B, the charge will remain as before and for A, the potential difference will remain same.

Charge on each capacitor in the previous case will be CV.
∴ Energy stored in capacitor A in circuit case is

Question 6.
Consider two conducting spheres of radii R₁ and R₂ with R₁ > R₂. If the two are at the same potential, the larger sphere has more charge than the smaller sphere. State whether the charge density of the smaller sphere is more or less than that of the larger one. (NCERT Exemplar)
Answer:

Question 7.
Calculate potential on the axis of a ring due to charge Q uniformly distributed along the ring of radius R.
(NCERT Exemplar)
Answer:
Let us take point P to be at a distance x from the centre of the ring, as shown in figure. The charge element dq is at a distance x from point P. Therefore, V can be written as

Long answer type questions

Question 1.
(i) Explain using suitable diagrams, the difference in the behaviour of a
(a) conductor and
(b) dielectric in the presence of external electric field. Define the terms polarisation of a dielectric and write its relation with susceptibility.

(ii) A thin metallic spherical shell of radius R carries a charge Q on its surface. A point charge \(\frac{Q}{2}\) is placed at its centre C and an other charge +2Q is placed outside the shell at a distance x from the centre as shown in the figure. Find (a) the force on the charge at the centre of shell and at the point A, (b) the electric flux through the shell.

Answer:
(i) When a capacitor is placed in an external electric field, the free charges present inside the conductor redistribute themselves in such a manner that the electric field due to induced charges opposes the external field within the conductor. This happens until a static- situation is achieved, i.e., when the two fields cancels each other and the net electrostatic field in the conductor becomes zero.

In contrast to conductors, dielectrics are non- conducting substances, i.e., they have no charge carriers. Thus, in a dielectric, free movement of charges is not possible. It turns out that the external field induces dipole moment by stretching molecules of the dielectric. The collective effect of all the molecular dipole moments is the net charge on the surface of the dielectric which produces a field that opposes the external field. However, the opposing field is so induced does not exactly cancel the external field, ft only reduces it. The extent of the effect depends on the nature of dielectric.

Both polar and non-polar dielectric develop net dipole moment in the presence of an external field. The dipole moment per unit volume is called polarisation and is denoted by P for linear isotropic dielectrics.
P = %E

(a) At point C, inside the shell.
The electric field inside a spherical shell is zero. Thus, the force experienced by charge at the centre C will also be zero.

Question 2.
(a) Find the ratio of the potential differences that must he applied across the parallel and series combination of two capacitors C ₁and C₂ with their capacitances in the ratio 1:2 so that the energy stored in the two cases becomes the same.

(b) Show that the potential energy of a dipole making angle θ with the direction of the field is given by u(θ) = – p \(\). Hence find out the amount of work done in rotating it from the position of unstable equilibrium to the stable equilibrium.

(b) As charges +q and -q traverse equal distance under equal and opposite forces; therefore, net work done in bringing the dipole in the region of electric field perpendicular to field-direction will be zero, i.e.,W₁ = 0.

Now, the dipole is rotated and brought to orientation making an angle θ with the field direction (i.e., θ₀ = 90° and θ₁ = θ°), therefore, work done
W₂ = p_E (cosθ₀ – cosθ ₁)
= p_E (cos 90° – cos θ) = -pE cos θ

∴Total work done in bringing the electric dipole from infinity, i.e., electric potential energy of electric dipole. Thus, work done by external torque in rotating a dipole in uniform electric field is stored as the potential energy of the system.

U = W₁ + W₂ = 0 – pEcosθ
= -pE cosθ
In vector form
U = – \(\vec{p} \cdot \vec{E}\)
For rotating dipole from position of unstable equilibrium (θ₀ =180°) to the stable equilibrium (θ = 0°)
∴W = pE (cos 180° – cos0°)
= pE (-1 -1) = -2 pE