JEE Advanced · 2017 · 21 May 2017 · Paper 1 · Geometrical Optics

A monochromatic light is travelling in a medium of refractive index

n = 1.6

. It enters a stack of glass layers from the bottom side at an angle

θ = 3 0^{\circ}

. The interfaces of the glass layers are parallel to each other. The refractive indices of different glass layers are monotonically decreasing as

n_{m} = n - m Δ n

, where

n_{m}

is the refractive index of the

m^{th}

slab and

Δ n = 0.1

(see the figure). The ray is refracted out parallel to the interface between the

(m - 1)^{th}

and

m^{th}

slabs from the right side of the stack. What is the value of

m

Answer: 8

Solution

When the ray emerges parallel to the interface, the angle of refraction in the

m

-th slab is

9 0^{\circ}

, meaning the ray undergoes total internal reflection at the critical angle. Applying Snell's law at the first interface:

n sin θ = n_{1} sin θ_{1}

, and continuing through the layers, we have

n sin θ = n_{m} sin 9 0^{\circ} = n_{m}

. Thus

n_{m} = n sin θ = 1.6 \times sin 3 0^{\circ} = 1.6 \times 0.5 = 0.8

. Since

n_{m} = n - m Δ n = 1.6 - 0.1 m

, we have

0.8 = 1.6 - 0.1 m

, giving

0.1 m = 0.8

, so

m = 8

JEE Advanced · 2017 · 21 May 2017 · Paper 1 · Integer answer