(a) Why does unpolarized light from a source show a variation in intensity when viewed through a Polaroid which is rotated? Show with the help of a diagram, how unpolarized light from sun gets linearly polarized by scattering.
(b) Three identical polarized sheets P1, P2 and P3 are oriented so that the pass axis of P2 and P3 are inclined at angles of 60°and 90° respectively with the pass axis of P1. A monochromatic source S of unpolarized light of intensity I˳ is kept in front of the polaroid sheet P1 as shown in figure. Determine the intensities of light as observed by the observer at O, when Polaroid P3 is rotated with respect to P2 at angles θ = 30° and 60°.

OR
(a) Derive an expression for path difference in Young’s double slit experiment and obtain the conditions for constructive and destructive interference at a point on the screen
(b) The intensity at the central maxima in Young’s double slit experiment is I˳. find out the intensity at a point where the path difference is ![]()
(a) A polaroid consists of long chain molecules aligned in a particular direction which gets absorbed.
Thus, if an un-polarized light wave is incident on such a Polaroid, it will get linearly polarized with the electric vector oscillating along a direction known as the pass-axis of the Polaroid, perpendicular to the aligned molecules.
So, its intensity is reduced by half. Rotating the Polaroid has no effect on the transmitted beam, and the transmitted intensity remains constant.
Polarization of sunlight by scattering:

(b)

The ray of light passing through Polaroid P1 will have intensity reduced by half. ![]()
Now, the Polaroid P2 is oriented at an angle 60° with respect to P1.
∴![]()
Now the Polaroid P3 is originally oriented at an angle 90- 60= 30°. Angle between P2 and P3 is 60°.
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Similarly, when P3 is rotated by 60°, angle between P2 and P3 is 90° ![]()
OR
(a)The diagrammatical representation of Young’s double slit experiment is:

S1 and S2 are two narrow slits illuminated by monochromatic light of wavelength λ on the screen XY.
IfS1 and S2 emit in same phase, then at O, superposition will be constructive and point will have central maxima.
The intensity at any point P depends on path difference i.e. S2P-S1P.
From geometry, we get,
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And ![]()
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As S1 and S2 are narrow slits.
∴![]()
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Condition for constructive interference: Phase difference between two superposing waves should be even multiple of π or path difference should be integral multiple of λ.
Condition for destructive interference: Phase difference between two superposing waves should be odd multiple of π or path difference should be integral multiple of λ/2.
The intensity of central maxima is I˳.
(b) Let I1 and I2 be the intensities emitted by S1 and S2.
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For central maxima: ![]()
Let I1= I2
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For path difference ![]()
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For path difference ![]()
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For path difference ![]()
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