A transparent cube of side d , made of a material of refractive index $\mu_{2}$, is immersed in a liquid of refractive index $\mu_{1}\left(\mu_{1}<\mu_{2}\right)$. A ray is incident on the face $A B$ at an angle $q$ (shown in the figure). Total internal reflection takes place at point E on the face BC . The q must satisfy :

(D) $\theta<\sin ^{-1} \sqrt{\frac{\mu_{2}^{2}}{\mu_{1}^{2}}-1}$

(A) $\theta>\sin ^{-1} \frac{\mu_{1}}{\mu_{2}}$

(B) $\theta>\sin ^{-1} \sqrt{\frac{\mu_{2}^{2}}{\mu_{1}^{2}}-1}$

(C) $\theta<\sin ^{-1} \frac{\mu_{1}}{\mu_{2}}$

JEE Main · 2019 · 12 Apr 2019 · Shift 2 · Geometrical Optics

A transparent cube of side d , made of a material of refractive index

μ_{2}

, is immersed in a liquid of refractive index

μ_{1} (μ_{1} < μ_{2})

. A ray is incident on the face

A B

at an angle

q

(shown in the figure). Total internal reflection takes place at point E on the face BC .

The q must satisfy :

Answer: (D)

Light entering face

A B

refracts according to

μ_{1} sin θ = μ_{2} sin r

. At

E

on the perpendicular face

B C

the angle of incidence is

9 0^{\circ} - r

, and total internal reflection needs this to exceed the critical angle

C

, where

sin C = \frac{μ _{1}}{μ _{2}}

. So

9 0^{\circ} - r > C

, that is

sin r < cos C = 1 - \frac{μ _{1}^{2}}{μ _{2}^{2}}

. Replacing

sin r = \frac{μ _{1}}{μ _{2}} sin θ

gives

\frac{μ _{1}}{μ _{2}} sin θ < \frac{μ _{2}^{2} - μ _{1}^{2}}{μ _{2}}

, so

sin θ < \frac{μ _{2}^{2}}{μ _{1}^{2}} - 1

and therefore

θ < sin^{- 1} \frac{μ _{2}^{2}}{μ _{1}^{2}} - 1

JEE Main · 2019 · 12 Apr 2019 · Shift 2 · Single correct option