$(a)\;\sin^{-1} \frac{\sqrt 3}{2} \\ (b)\;\sin ^{-1} \frac{\sqrt 3 }{4} \\ (c)\;\sin ^{-1} \frac{\sqrt 3}{2 \sqrt 2} \\ (d)\;\sin ^{-1} \frac{1}{\sqrt 2} $

Since the incident light is in rarer medium.

Total internal reflection cannot take place.

$C= \sin ^{-1} \large\frac{1}{\mu}$$=30^{\circ}$

$1=2C=60^{\circ}$

Applying Snell's law $1 \sin 60^{\circ}=2 \sin r$

$\sin r =\large\frac{\sqrt 3}{4}$

$r= \sin^{-1} \bigg(\large\frac{\sqrt 3}{4}\bigg)$

Hence b is the correct answer.

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