In the figure,suppose the transmission axes o | vedclass

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(A) $1$. When unpolarized light of intensity $I_{max}$ passes through the first polarizer,the intensity of the transmitted light is $I_1 = \frac{I_{ma

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

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(A) $1$. When unpolarized light of intensity $I_{max}$ passes through the first polarizer,the intensity of the transmitted light is $I_1 = \frac{I_{max}}{2}$.$2$. The second polarizer is rotated at an angular speed $\omega$,so the angle between the transmission axes of the first and second polarizers is $	heta = \omega t$. According to Malus's Law,the intensity of light emerging from the second polarizer is $I_2 = I_1 \cos^2 	heta = \frac{I_{max}}{2} \cos^2 \omega t$.$3$. The third polarizer is fixed at an angle of $90^\circ$ with respect to the first polarizer. Therefore,the angle between the transmission axes of the second and third polarizers is $(90^\circ - 	heta) = (90^\circ - \omega t)$.$4$. Applying Malus's Law again for the third polarizer: $I = I_2 \cos^2(90^\circ - \omega t) = I_2 \sin^2 \omega t$.$5$. Substituting $I_2$: $I = (\frac{I_{max}}{2} \cos^2 \omega t) \sin^2 \omega t = \frac{I_{max}}{2} (\sin \omega t \cos \omega t)^2 = \frac{I_{max}}{2} (\frac{\sin 2\omega t}{2})^2 = \frac{I_{max}}{8} \sin^2 2\omega t$.$6$. Using the identity $\sin^2 \phi = \frac{1 - \cos 2\phi}{2}$,we get $I = \frac{I_{max}}{8} \cdot \frac{1 - \cos 4\omega t}{2} = \frac{I_{max}}{16} (1 - \cos 4\omega t)$.

$A$. Angle of reflection	$(i)$ $2 \sin^{-1} \left(\sqrt{\frac{2}{3}}\right)$
$B$. Angle of refraction	$(ii)$ $\sin^{-1} \left(\sqrt{\frac{2}{3}}\right) - \sin^{-1} \left(\frac{1}{\sqrt{3}}\right)$
$C$. Angle between incident and completely polarised light	$(iii)$ $\sin^{-1} \left(\frac{1}{\sqrt{3}}\right)$
$D$. Angle of deviation of the incident ray	$(iv)$ $\cos^{-1} \left(\frac{1}{\sqrt{3}}\right)$

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