JEE Advanced · Physics · 24. Ray Optics
A light beam is travelling from Region I to Region IV (refer figure). The refractive index in Region I, II, III and IV are \(n_0, \frac{n_0}{2}, \frac{n_0}{6}\) and \(\frac{n_0}{8}\), respectively. The angle of incidence \(\theta\) for which the beam just misses entering Region IV is
- A \(\sin ^{-1}\left(\frac{3}{4}\right)\)
- B \(\sin ^{-1}\left(\frac{1}{8}\right)\)
- C \(\sin ^{-1}\left(\frac{1}{4}\right)\)
- D \(\sin ^{-1}\left(\frac{1}{3}\right)\)
Answer & Solution
Correct Answer
(B) \(\sin ^{-1}\left(\frac{1}{8}\right)\)
Step-by-step Solution
Detailed explanation
Critical angle from region III to region IV
\(\sin \theta_c=\frac{n_0 / 8}{n_0 / 6}=\frac{3}{4}\)
Now applying Snell's law in region I and region III :
\(n_0 \sin \theta=\frac{n_0}{6} \sin \theta_c\)
or
\(\begin{aligned}\sin \theta & =\frac{1}{6} \sin \theta_c=\frac{1}{6}\left(\frac{3}{4}\right)=\frac{1}{8} \\\theta & =\sin ^{-1}\left(\frac{1}{8}\right)\end{aligned}\)
\(\therefore\) correct option is (b).
\(\sin \theta_c=\frac{n_0 / 8}{n_0 / 6}=\frac{3}{4}\)
Now applying Snell's law in region I and region III :
\(n_0 \sin \theta=\frac{n_0}{6} \sin \theta_c\)
or
\(\begin{aligned}\sin \theta & =\frac{1}{6} \sin \theta_c=\frac{1}{6}\left(\frac{3}{4}\right)=\frac{1}{8} \\\theta & =\sin ^{-1}\left(\frac{1}{8}\right)\end{aligned}\)
\(\therefore\) correct option is (b).
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