Two polaroids are placed in the path of an un | vedclass

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(A) Let the initial intensity of the unpolarized light be $I_0$.When unpolarized light passes through the first polaroid,the intensity of the transmit

Vedclass · Accepted Answer

$\left( \frac{I_0}{8} ight) \sin^2 2	heta$

Vedclass · Answer

(A) Let the initial intensity of the unpolarized light be $I_0$.When unpolarized light passes through the first polaroid,the intensity of the transmitted light is $I_1 = \frac{I_0}{2}$.The third polaroid is placed at an angle $	heta$ with respect to the first polaroid. According to Malus's Law,the intensity of light emerging from the third polaroid is $I_2 = I_1 \cos^2 	heta = \frac{I_0}{2} \cos^2 	heta$.The second polaroid is initially at $90^\circ$ to the first. Since the third polaroid is at $	heta$ to the first,the angle between the third and the second polaroid is $(90^\circ - 	heta)$.Applying Malus's Law again for the light passing through the second polaroid:$I_3 = I_2 \cos^2(90^\circ - 	heta) = I_2 \sin^2 	heta$.Substituting the value of $I_2$:$I_3 = \left( \frac{I_0}{2} \cos^2 	heta ight) \sin^2 	heta = \frac{I_0}{2} (\sin 	heta \cos 	heta)^2$.Using the identity $\sin 2	heta = 2 \sin 	heta \cos 	heta$,we have $\sin 	heta \cos 	heta = \frac{\sin 2	heta}{2}$.Therefore,$I_3 = \frac{I_0}{2} \left( \frac{\sin 2	heta}{2} ight)^2 = \frac{I_0}{2} \cdot \frac{\sin^2 2	heta}{4} = \frac{I_0}{8} \sin^2 2	heta$.

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