Q. A beam of light is travelling from region II to region III (see the figure). The refractive index in region I, II and III are$n_{0},\frac{n_{0}}{\sqrt{2}},\ $and$\frac{n_{0}}{2}$respectively. The angle of incidence $\theta$ for which the beam just misses entering region III is
A) $30{^\circ}$
B) $45{^\circ}$
C) $60{^\circ}$
D) $\sin^{- 1}\left( \sqrt{2} \right)$
Explanation
(a) The critical angle for region II and III
$\sin{C = \frac{\mu_{III}}{\mu_{II}}} = \frac{\frac{n_{0}}{2}}{\frac{n_{0}}{\sqrt{2}}} = \frac{1}{\sqrt{2}}$
$ie.\ \angle\ C = 45{^\circ}$
The ray, if incident at $45{^\circ}$ at the interface of II and III it will be totally internally reflected.
Now, from Snell’s lay in region I and II.
$n_{0}\sin\theta = \frac{n_{0}}{\sqrt{2}}\sin C$
$Or\sin\theta = \frac{1}{\sqrt{2}} \times \sin{45{^\circ}}$
$Or\sin\theta = \frac{1}{\sqrt{2}} \times \frac{1}{\sqrt{2}}$
$Or\sin\theta = \frac{1}{2}$
$Or\ \theta = 30{^\circ}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.