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(A) Let $I_0$ be the initial intensity of unpolarized light.After passing through the polarizer,the intensity becomes $I_p = \frac{I_0}{2}$.According

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(A) Let $I_0$ be the initial intensity of unpolarized light.After passing through the polarizer,the intensity becomes $I_p = \frac{I_0}{2}$.According to Malus' Law,the intensity $I$ emerging from the analyser is $I = I_p \cos^2 	heta$,where $	heta$ is the angle between the transmission axes of the polarizer and the analyser.Given $I = \frac{I_0}{10}$,we have $\frac{I_0}{10} = \frac{I_0}{2} \cos^2 	heta$.$\cos^2 	heta = \frac{1}{5} \implies \cos 	heta = \frac{1}{\sqrt{5}} \approx 0.447$.$	heta = \cos^{-1}(0.447) \approx 63.43^o$.To reduce the output intensity to zero,the analyser must be rotated until the angle between the axes is $90^o$.The additional angle required is $\Delta 	heta = 90^o - 63.43^o = 26.57^o$.

$A$. 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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