A light beam is travelling 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

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

Question

A light beam is travelling 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

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

Toppr Admin · Accepted Answer

Total internal reflection occurs at the interface of region III and IV as shown in figure-n0sin-x3B8-no2sin-x3B8-1-no6sin-x3B8-2-no8sinrn0sin-x3B8-n08sin900-x21D2-x3B8-sin-x2212-1-18-

A light beam is travelling from Region I to Region IV (Refer 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 issin−1(34)sin−1(18)sin−1(14)sin−1(13)

Total internal reflection occurs at the interface of region III and IV as shown in figure.n0sinθ=no2sinθ1=no6sinθ2=no8sinrn0sinθ=n08sin900⇒θ=sin−1(18)

Total internal reflection occurs at the interface of region III and IV as shown in figure.

$n_{0} sin θ = \frac{n_{o}}{2} sin θ_{1} = \frac{n_{o}}{6} sin θ_{2} = \frac{n_{o}}{8} sin r$

$n_{0} sin θ = \frac{n_{0}}{8} sin 90^{0} \Rightarrow θ = {sin}^{- 1} (\frac{1}{8})$