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Q.
The critical angle of a prism is $30^{\circ}$. The velocity of light in the medium is :
J & K CETJ & K CET 2001
Solution:
The critical angle is the angle of incidence in the denser medium for which the angle of refraction in the rarer medium is $90^{\circ}$.
$\therefore \mu=\frac{\sin 90^{\circ}}{\sin C}=\frac{1}{\sin C}$ ... (i)
Also, $\mu=\frac{v_{0}}{v}$...(ii)
where $v_{0}$ is speed of light in vacuum and $v$ is speed in medium.
From Eqs. (i) and (ii), we get
$v=v_{0} \sin C=3 \times 10^{8} \times \frac{1}{2}$
$=1.5 \times 10^{8} m / s$