A light beam is travelling from Region I to IV (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 [2008]

Question

Smart Achievers · Accepted Answer

(2)

For refraction at parallel interfaces,

From Snell's law,

$n_{0} \sin θ = \frac{n_{0}}{2} \sin α = \frac{n_{0}}{6} \sin β = \frac{n_{0}}{8} \sin 90 °$

The angle of refraction in region IV must be $90 °$ as the beam just misses entering region IV.

$∴$ $\sin θ = \frac{1}{8}$

or, $θ = \sin^{- 1} (\frac{1}{8})$

Solution

Q.
A light beam is travelling from Region I to IV (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 [2008]

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

2 $\sin^{- 1} (\frac{1}{8})$

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

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

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Solution

Q. A light beam is travelling from Region I to IV (figure). The refractive index in regions I, II, III and IV are n0, n02, n06 and n08 respectively. The angle of incidence θ for which the beam just misses entering region IV is [2008]

1 sin-1(34)

2 sin-1(18)

3 sin-1(14)

4 sin-1(13)