In the figure,suppose the transmission axes o | vedclass

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(A) According to Malus's law,the intensity of light transmitted through a polarizer is $I = I_0 \cos^2 	heta$,where $	heta$ is the angle between the

Vedclass · Accepted Answer

$I=\frac{1}{16}I_{max}(1-\cos 4\omega t)$

Vedclass · Answer

(A) According to Malus's law,the intensity of light transmitted through a polarizer is $I = I_0 \cos^2 	heta$,where $	heta$ is the angle between the transmission axes.$1$. When unpolarized light of intensity $I_{max}$ passes through the first (left) polarizer,the intensity becomes $I_1 = \frac{I_{max}}{2}$.$2$. The second (center) polarizer is at an angle $	heta = \omega t$ with respect to the first. The intensity after the second polarizer is $I_2 = I_1 \cos^2 	heta = \frac{I_{max}}{2} \cos^2(\omega t)$.$3$. The third (right) polarizer is at an angle $(90^\circ - 	heta)$ with respect to the second polarizer. The intensity after the third polarizer is $I = I_2 \cos^2(90^\circ - 	heta) = I_2 \sin^2 	heta$.$4$. Substituting $I_2$: $I = \left(\frac{I_{max}}{2} \cos^2 	hetaight) \sin^2 	heta = \frac{I_{max}}{2} (\sin 	heta \cos 	heta)^2 = \frac{I_{max}}{2} \left(\frac{\sin 2	heta}{2}ight)^2 = \frac{I_{max}}{8} \sin^2 2	heta$.$5$. Using the identity $\sin^2 	heta = \frac{1 - \cos 2	heta}{2}$,we get $I = \frac{I_{max}}{8} \left(\frac{1 - \cos 4	heta}{2}ight) = \frac{I_{max}}{16} (1 - \cos 4\omega t)$.

$X$	$1$	$2$	$3$	$4$	$5$
$P(X)$	$k^2$	$2k$	$k$	$2k$	$5k^2$

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