CBSE Class 12 Physics 2017 Delhi Set 1 Paper

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Question : 25
Total: 26
(a) Distinguish between unpolarised light and linearly polarized light. How does one get linearly polarised light with the help of a polaroid?
(b) A narrow beam of unpolarised light of intensity I0 is incident on a polaroid P1. The light transmitted by it is then incident on a second polaroid P2 with its pass axis making angle of 60 relative to the pass axis of P1. Find the intensity of the light transmitted by P2.
OR
(a) Explain two features to distinguish between the interference pattern in Young's double slit experiment with the diffraction pattern obtained due to a single slit.
(b) A monochromatic light of wavelength 500nm is incident normally on a single slit of width 0.2mm to produce a diffraction pattern. Find the angular width of the central maximum obtained on the screen.
Estimate the number of fringes obtained in Young's double slit experiment with angular fringe width 0.5 , which can be accommodated within the region of total angular spread of the central maximum due to single slit.
Solution:  
(a) Distinction between unpolarised and linearly polarized light 2 Obtaining linearly polarized
(b) Calculation of intensity of light
(a) In an unpolarised light, the oscillations of the electric field, are in random directions in planes perpendicular to the direction of propagation. For a polarized light, the oscillations are aligned along one particular direction.
Alternatively
Polarized light can be distinguished from unpolarised light when it is allowed to pass through a polaroid. Polarized light can show change in its intensity, on passing through a Polaroid; intensity remains same in case of unpolarised light.
When unpolarised light wave is incident on a polaroid, then the electric vectors along the direction of its aligned molecules, get absorbed; the electric vector, oscillating along a direction perpendicular to the aligned molecules, pass through. Thislight is called linearlv polarized light.
(b) According to Malus' Law:
I=I0cos2θ
I=(
I0
2
)
cos2
θ
,

where I0 is the intensity of unpolarised light.
θ=60 (given)
I=
I0
2
cos2
60
=
I0
2
×(
1
2
)
2

=
I0
8

OR
(a) Explanation of two features (distinguishing between interference pattern and diffraction pattern.)
(b) Calculation of angular width of central maxima
Estimation of number of fringes
(a)
 Interference Pattern  Diffraction pattern
 (i) All fringes are of equal width.  (i) Width of central maxima is twice the width of higher order bands.
 (ii) Intensity of all bright bands is equal.  (ii) Intensity goes on decreasing for higher order of diffraction bands.
(b) Angular width of central maximum
ω=
2λ
a

=
2×500×109
0.2×103
radian

=5×103 radian
β=
λD
d

Linear width of central maxima in the diffraction pattern
ω=
2λD
a

Let ' n ' be the number of interference fringes which can be accommodated in the central maxima
n×β=ω
n=
2λD
a
×
d
λD

n=
2d
a

[Note : Award the last 12 mark if the student writes the answers as 2 (taking d=a ), or just attempts to do these calculation.]
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