JEE Advanced · 2008 · 13 Apr 2008 · Paper 2 · Geometrical Optics

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A light beam is traveling from Region I to Region IV (Refer Figure). The refractive index in Regions 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

θ

for which the beam just misses entering Region IV is

(A) $sin^{- 1} (\frac{3}{4})$
(B) $sin^{- 1} (\frac{1}{8})$ ✓ Correct
(C) $sin^{- 1} (\frac{1}{4})$
(D) $sin^{- 1} (\frac{1}{3})$

Answer: (B)

Solution

For the beam to just miss entering Region IV, total internal reflection must occur at the interface between Region III and Region IV.

At the critical angle at III-IV interface:

\frac{n _{0}}{6} sin (90°) = \frac{n _{0}}{8} sin (90°)

This is not the right approach. Since the interfaces are parallel, by Snell's law applied from Region I to Region III:

n_{0} sin θ = \frac{n _{0}}{6} sin θ_{I I I}

For total internal reflection at III-IV boundary:

\frac{n _{0}}{6} sin (90°) = \frac{n _{0}}{8} \times 1

But this gives the critical angle at that interface. Using the parallel interface property:

n_{0} sin θ = \frac{n _{0}}{8} sin (90°)

sin θ = \frac{1}{8}

JEE Advanced · 2008 · 13 Apr 2008 · Paper 2 · Single correct option

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JEE Advanced · 2008 · 13 Apr 2008 · Paper 2 · Geometrical Optics

Solution

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