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Question

(C) Let the intensity of light after passing through the first polaroid be $I_0$.When a second polaroid is placed at an angle $	heta = 22.5^{\circ}$

Vedclass · Accepted Answer

$\frac{I_0}{8}$

Vedclass · Answer

(C) Let the intensity of light after passing through the first polaroid be $I_0$.When a second polaroid is placed at an angle $	heta = 22.5^{\circ}$ with the first,the intensity of light transmitted through it is given by Malus' Law: $I_1 = I_0 \cos^2 	heta$.The third polaroid is placed at an angle $(90^{\circ} - 	heta)$ with respect to the second polaroid.The intensity of light transmitted through the third polaroid is $I_2 = I_1 \cos^2(90^{\circ} - 	heta) = I_0 \cos^2 	heta \sin^2 	heta$.Using the identity $\sin(2	heta) = 2 \sin 	heta \cos 	heta$,we have $\sin^2(2	heta) = 4 \sin^2 	heta \cos^2 	heta$,so $\sin^2 	heta \cos^2 	heta = \frac{\sin^2(2	heta)}{4}$.Substituting this into the expression for $I_2$: $I_2 = I_0 \frac{\sin^2(2	heta)}{4}$.Given $	heta = 22.5^{\circ}$,then $2	heta = 45^{\circ}$.$I_2 = \frac{I_0}{4} \sin^2(45^{\circ}) = \frac{I_0}{4} 	imes (\frac{1}{\sqrt{2}})^2 = \frac{I_0}{4} 	imes \frac{1}{2} = \frac{I_0}{8}$.

The intensity of transmitted light when a polaroid sheet is placed between two crossed polaroids at $22.5^{\circ}$ from the polarization axis of one of the polaroids is (where $I_0$ is the intensity of polarized light after passing through the first polaroid):

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