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A right angle prism $(45^{\circ} - 90^{\circ}- 45^{\circ})$ of refractive index n has a plate of refractive index $n_1\;( n_1 < n ) $ to its diagonal wall. The assembly is in air. A ray is incident at AB . Then the angle of incidence at AB for which the ray strikes the diagonal face at critical angle is :-

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

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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.
answered Feb 20, 2014 by meena.p

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