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Geometrical Optics

Section: Physics

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Question : 55 of 56

Marks: +1, -0

A light beam is travelling from Region I to Region IV (refer to the figure). The refractive index in Regions I, II, III and IV are

n_0, \frac{n_0}{6}, \frac{n_0}{6}

\frac{n_0}{8}

, respectively. The angle of incidence θ for which the beam just misses entering Region IV is

[JEE Adv 2008 P2]

$\sin^{-1}\left(\frac{3}{4}\right)$
$\sin^{-1}\left(\frac{1}{8}\right)$
$\sin^{-1}\left(\frac{1}{4}\right)$
$\sin^{-1}\left(\frac{1}{3}\right)$

Solution:

Let i be the angle of incidence at the interface of Region III and Region IV. Using shell’s law for Region I and Region III, we get

n_0\sin\theta

=

\frac{n_0}{6} \sin i

For a total internal reflection in Region III, we have

i ≥

i_C

(Sin i) ≥

\left(\frac{\frac{n_0}{8}}{\frac{n_0}{6}}\right)

(sin i) ≥

\left(\frac{3}{4}\right)

Therefore,

n_0 \sin \theta

=

\frac{n_0}{6} \times \frac{3}{4}

=

\frac{n_0}{8}

⇒ sin θ =

\frac{1}{8}

⇒ θ =

\sin^{-1}\left(\frac{1}{8}\right)

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