Monochromatic light is incident on a glass prism of angle A. If the refractive index of the material of the prism is $\mu$, a ray, incident at an angle $\theta$, on the face AB would get transmitted through the face AC of the prism provided:

(D) $\theta>\sin ^{-1}\left[\mu \sin \left(\mathrm{~A}-\sin ^{-1}\left(\frac{1}{\mu}\right)\right)\right]$

(A) $\theta<\sin ^{-1}\left[\mu \sin \left(\mathrm{~A}-\sin ^{-1}\left(\frac{1}{\mu}\right)\right)\right]$

(B) $\theta>\cos ^{-1}\left[\mu \sin \left(A+\sin ^{-1}\left(\frac{1}{\mu}\right)\right)\right]$

(C) $\theta<\cos ^{-1}\left[\mu \sin \left(A+\sin ^{-1}\left(\frac{1}{\mu}\right)\right)\right]$

JEE Main · 2015 · 04 Apr 2015 · Offline · Geometrical Optics

Monochromatic light is incident on a glass prism of angle A. If the refractive index of the material of the prism is

μ

, a ray, incident at an angle

θ

, on the face AB would get transmitted through the face AC of the prism provided:

Answer: (D)

Refraction at face

A B

gives

sin θ = μ sin r

. To escape through

A C

rather than suffer total internal reflection there, the ray's angle on

A C

must stay below the critical angle

θ_{C} = sin^{- 1} (1/ μ)

. The prism geometry fixes

r + r^{'} = A

, so

r^{'} < θ_{C}

means

r > A - sin^{- 1} (1/ μ)

. Taking sines and multiplying by

μ

sin θ = μ sin r > μ sin (A - sin^{- 1} (1/ μ))

, so transmission needs

θ > sin^{- 1} [μ sin (A - sin^{- 1} (1/ μ))]

JEE Main · 2015 · 04 Apr 2015 · Offline · Single correct option