$(a)\;\sin ^{-1} \frac{1}{\sqrt 2} \bigg[n^2-n_1^2)^{1/2}-n_1\bigg] \\ (b)\;\sin ^{-1} \frac{1}{\sqrt 2} \bigg[\sqrt {n^2-n_1^2}\bigg] \\ (c)\;\sin ^{-1} \bigg[\sqrt {n^2-n_1^2}-n_1\bigg] \\ (d)\;None$

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Let the critical angle at face AC be $\theta_c$, At face AB from snell's law, $|\sin i|=n \sin r$

But from geometry in the triangle ADE, we have,

$(90- \theta_c)+(90-r)+45=180$

ie $r=(45-\theta_c)$

=> $1 \sin i =n \sin (45 - \theta_c) =\large\frac{n}{\sqrt 2}$$ [ \cos \theta_c- \sin \theta_c]$

But as here $\mu =\frac{n}{n_1}$

So $sin \theta_c=\bigg[\large\frac{1}{\mu}\bigg] =\frac{n_1}{n}$

and $ \cos \theta_c=\sqrt {1- \sin ^2 \theta_c}$

$\qquad= \large\frac{1}{n} $$ \sqrt {n^2-n_1^2}$

So equation becomes

$\sin i =\large\frac{1}{\sqrt 2} $$\bigg[(n^2-n_1^2)^{1/2} -n_1 \bigg]$

ie $i= \sin ^{-1} \frac{1}{\sqrt 2} \bigg[n^2-n_1^2)^{1/2}-n_1\bigg]$

Hence a is the correct answer.

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