When a polaroid sheet is rotated between two | vedclass

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(D) Let $I_0$ be the intensity of unpolarized light incident on the first polaroid.$I_1 = I_0 / 2$ is the intensity of light transmitted from the firs

Vedclass · Accepted Answer

$45$

Vedclass · Answer

(D) Let $I_0$ be the intensity of unpolarized light incident on the first polaroid.$I_1 = I_0 / 2$ is the intensity of light transmitted from the first polaroid.Let $	heta$ be the angle between the first and second polaroid,and $\phi$ be the angle between the second and third polaroid.Since the first and third polaroids are crossed,$	heta + \phi = 90^{\circ}$,so $\phi = 90^{\circ} - 	heta$.The intensity transmitted from the second polaroid is $I_2 = I_1 \cos^2 	heta$.The intensity transmitted from the third polaroid is $I_3 = I_2 \cos^2 \phi = I_1 \cos^2 	heta \cos^2 (90^{\circ} - 	heta) = I_1 \cos^2 	heta \sin^2 	heta$.Using the identity $\sin 2	heta = 2 \sin 	heta \cos 	heta$,we get $I_3 = I_1 (\sin 2	heta / 2)^2 = (I_0 / 2) \cdot (\sin^2 2	heta / 4) = (I_0 / 8) \sin^2 2	heta$.$I_3$ is maximum when $\sin^2 2	heta = 1$,which implies $2	heta = 90^{\circ}$,so $	heta = 45^{\circ}$.

When a polaroid sheet is rotated between two crossed polaroids,then the transmitted light intensity will be maximum for a rotation of: (in $^{\circ}$)

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