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# Find angle of refraction in a medium $(\mu =2)$ . If light is incident in vacuum making angle equal to twice the critical angle.

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

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.