Question

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

• A. $56.4^{\circ }$
• B. $28.2^{\circ }$
• C. $50.4^{\circ }$
• D. $39.4^{\circ }$

Answer 1

Answer: (A)

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, $v_d=2\times 10^{8}m/s$,
$v_r=2.4\times 10^{8}m/s$
$\therefore \sin C=\frac{2\times 10^8}{2.4\times 10^8}=\frac{5}{6}$
$\Rightarrow C={\sin }^{-1}\left(\frac{5}{6}\right)=56.4^{\circ }$