﻿ (a) (i) Two independent monochromatic sources of light cannot produce a sustained interference pattern . Give reason. (ii) Light waves each of amplitude "a" and frequency "ω", emanating from two coherent light sources superpose at a point. If the displacements due to these waves is given by y1 = a cos ωt and y2 = a cos(ωt + ϕ) where ϕ is the phase difference between the two, obtain the expression for the resultant intensity at the point.(b) In Young s double slit experiment, using monochromatic light of wavelength λ, the intensity of light at a point on the screen where path difference is λ, is K units. Find out the intensity of light at a point where path difference is λ/3. from Class 12 CBSE Previous Year Board Papers | Physics 2014 Solved Board Papers

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# CBSE Class 12 Physics Solved Question Paper 2014

#### Long Answer Type

31.

(a) How does one demonstrate, using a suitable diagram, that unpolarized light when passed through a Polaroid gets polarized?

(b) A beam of unpolarized light is incident on a glass-air interface. Show, using a suitable ray diagram, that light reflected from the interface is totally polarized, when μ = tan iB, where μ is the refractive index of glass with respect to air and iB is the Brewster's angle. Polarization of light is referred to as restricting the vibration of light in a perpendicular direction perpendicular to the direction of propagation of wave.

The vibration of particles of light which is parallel to the axis of crystal passes through the Polaroid on passing an unpolarized light. All other vibrations are absorbed and that is why intensity of emerging light is reduced.

The plane of vibration here is ABCD, in which the vibrations of the polarized light is confined and the plane KLMN is called plane of polarization. KLMN is perpendicular to the plane of vibration. Reflected light is totally polarized, when unpolarized light is incident on the glass-air interface at the Brewster angle iB. This is known as Brewster’s law.

The reflected component OB and refracted component OC are mutually perpendicular to each other, when light is incident at Brewster’s angle. Now, using Snell’s law, we have i = iB and r = (900 – iB)

Therefore, 1239 Views

# 32.(a) (i) 'Two independent monochromatic sources of light cannot produce a sustained interference pattern'. Give reason.(ii) Light waves each of amplitude "a" and frequency "ω", emanating from two coherent light sources superpose at a point. If the displacements due to these waves is given by y1 = a cos ωt and y2 = a cos(ωt + ϕ) where ϕ is the phase difference between the two, obtain the expression for the resultant intensity at the point.(b) In Young's double slit experiment, using monochromatic light of wavelength λ, the intensity of light at a point on the screen where path difference is λ, is K units. Find out the intensity of light at a point where path difference is λ/3.

(a)

(i) The condition for the sustained interference is that both the sources must be coherent (i.e. they must have the same wavelength and the same frequency, and they must have the same phase or constant phase difference).
Two sources are monochromatic if they have the same frequency and wavelength. Since they are independent, i.e. they have different phases with irregular difference, they are not coherent sources.

ii) Let the displacement of the waves from the sources S1 and S2 at point P on the screen at any time t be given by:

y1 = a cos ωt

y2 = a cos (ωt + Φ)

where, Φ is the constant phase difference between the two waves.

By the superposition principle, the resultant displacement at point P is given by:

y = y1 + y2

y = a cos ωt a cos (ωt + Φ)

=2 a[cos cos y = 2 acos cos ... (i) ... (2)

Then, equation (i) becomes:

y = A cos (ωt+ Now, we have: ... (3)

The intensity of light is directly proportional to the square of the amplitude of the wave. The intensity of light at point P on the screen is given by:

I = 4 a2 cos2 ( )                                  ... (4)

(b) Wavelength of monochromatic light = Path difference = .
So, phase difference, Intensity of light = K units

Intensity is given by, I = When path difference is Intensity of light, I’= =  2994 Views

33.

(a) State briefly the processes involved in the formation of p-n junction explaining clearly how the depletion region is formed.

(b) Using the necessary circuit diagrams, show how the V-I characteristics of a p-n junction are obtained in
(i) Forward biasing

(ii) Reverse biasing

How these characteristics are made use in rectification?

(a) The n-type semi-conductor has more concentration of electrons than that of a holes and p-type semi-conductor has more concentration of holes. Holes diffuse from p-side to n-side whereas electrons diffuse from n-side to p-side due to difference in concentration of charge carriers.

An ionized donor is left behind on n-side when electron diffuses from n side to p-side. The ionized donor (+ ve charge) is immobile as it is bound by the surrounding atoms. Therefore, a layer of positive charge is developed on the n-side of the junction. Similarly, a layer of negative charge is developed on the p-side. Hence, a space-charge region is formed on either side of the junction, which has immobile ions and is devoid of any charge carrier, called as depletion layer or depletion region. (b)

For a p-n junction diode under forward bias, p-side is connected to the positive terminal and n-side is connected to the negative terminal. When voltage is applied, electrons in n-region and holes in the p-region moves towards the p-n junction. Hence, there is decrease in the width of the depletion region thereby, offering less resistance. Diffusion of majority carriers takes place in the junction giving rise to a forward current.

The V-I characteristic of p-n junction in forward bias is shown below: (ii) When p-n junction diode is reverse biased, the positive terminal of battery is connected to n-side and negative terminal to p-side. The barrier height increases and the width of depletion region also increase as a result of reverse biasing. There is no conduction across the junction because of the lack of majority charge carriers. After applying a high reverse biased voltage, few minority carriers cross the junction. Hence, a current flows in reverse direction which is known as the reverse current.

The V-I characteristic of p-n junction diode in reverse bias is shown below: p-n junction can be used for rectification purpose. Its working is based on the fact that, resistance of junction becomes low when forward biased and R becomes high when reverse biased.

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34.

(a) Differentiate between three segments of a transistor on the basis of their size and level of doping.

(b) How is a transistor biased to be in active state?

(c) With the help of necessary circuit diagram, describe briefly how n-p-n transistor in CE configuration amplifies a small sinusoidal input voltage. Write the expression for the ac current gain.

On the basis of size and level of doping,

(a)

Emitter (E) - Emitter is heavily doped and is the left hand side thick layer of the transistor.

Base (B) - It is the central thin layer of the transistor, which is lightly doped.

Collector (C) - It is the right hand side thick layer of the transistor, which is moderately doped.

(b)

There are two conditions for a transistor to be into an active region.

1. The input circuit should be forward biased by using a low voltage battery.

2. The output circuit should be reverse biased by using a high voltage battery.

(c)

n-p-n Transistor as an amplifier:

The input circuit is forward biased by using a low voltage battery. Hence, the resistance of the input circuit is small. The output circuit is reverses biased using a high voltage battery. Hence, the resistance of the output circuit is large.

The operating point is fixed in the middle of its active region. The output is taken between the collector and the ground.

Applying Kirchhoff’s law to the output loop:

VCC = VCE + ICRC

If Vc is the collector voltage then,

Vc = VCE − ICRC                                                      ... (A)

When the input signal voltage is fed to the emitter base circuit, it will change the emitter voltage and hence to the emitter current, which in turn will change the collector current. Due to this the collector voltage VC will vary in accordance with relation (A). This variation in collector voltage appears as the amplified output.

VBB = VBE + IBRB

Here, vi ≠ O

Then, VBB + vi = VBE + IBRB + ΔIB (RB + ri)  ; is the required expression for current gain.

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