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Q.
Light travels in two medium $A$ and $B$ with speeds $2 \times 10^{8} m / s$ and $2.4 \times 10^{8} m / s$ respectively. The critical angle $C$ between them will be
AMUAMU 2000
Solution:
The relation between critical angle $(C)$ and refractive index $(\mu)$ is
$\sin C=\frac{1}{\mu}$...(i)
Also, $ \mu=\frac{v_{r}}{v_{d}}$ ... (ii)
From Eqs. (i) and (ii), we get
$\sin C=\frac{v_{d}}{v_{r}}$
Given, $v_{d}=2 \times 10^{8} m / s$,
$v_{r}=2.4 \times 10^{8} m / s$
$\therefore \sin C=\frac{2 \times 10^{8}}{2.4 \times 10^{8}}=\frac{5}{6}$
$\Rightarrow C=\sin ^{-1}\left(\frac{5}{6}\right)=56.4^{\circ}$