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← JEE Advanced 2017 · Paper 1
PhysicsInteger+3Question 10 of 54

JEE Advanced 2017 · Paper 1, Physics question 10

A monochromatic light is travelling in a medium of refractive index n=1.6n = 1.6. It enters a stack of glass layers from the bottom side at an angle θ=30∘\theta = 30^\circ. The interfaces of the glass layers are parallel to each other. The refractive indices of different glass layers are monotonically decreasing as nm=n−mΔnn_m = n - m\Delta n, where nmn_m is the refractive index of the mthm^{th} slab and Δn=0.1\Delta n = 0.1 (see the figure). The ray is refracted out parallel to the interface between the (m−1)th(m-1)^{th} and mthm^{th} slabs from the right side of the stack. What is the value of mm?

A stack of horizontal glass slabs numbered 1, 2, 3, …, m−1, m from the bottom, with refractive indices n − Δn, n − 2Δn, n − 3Δn, …, n − (m−1)Δn, n − mΔn; below the stack is the medium of index n. A ray enters from below at angle θ to the normal, bends away from the normal in each slab, and finally travels parallel to the interface between slabs m−1 and m, leaving the stack on the right.
Answer and worked solution

Official answer: 8

Invariant. With parallel interfaces, nsin⁡θn\sin\theta is the same in every layer:
nsin⁡θ=1.6×12=0.8.n\sin\theta=1.6\times\tfrac12=0.8.

Grazing exit. The ray emerges parallel to the interface between slabs m−1m-1 and mm, so it grazes in slab mm: nmsin⁡90∘=0.8n_m\sin90^\circ=0.8.

Solve.
1.6−0.1m=0.8  ⟹  m=8.1.6-0.1m=0.8\implies m=8.

Answer: 8