JEE Advance - Physics (2017 - Paper 1 Offline - No. 10)
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 $$\theta = {30^ \circ }.$$ The interfaces of the glass layers are parallel to each other. The refractive indices of different glass layers are monotonically decreasing as $${n_m} = n - m\Delta n,$$ where $${n_m}$$ is the refractive index of the $${m^{th}}$$ slab and $$\Delta n = 0.1$$ (see the figure). The ray is refracted out parallel to the interface between the $${\left( {m - 1} \right)^{th}}$$ and $${m^{th}}$$ slabs from the right side of the stack. What is the value of $$m$$?
Answer
8
Explanation
Applying Snell's law, we have nsin$$\theta$$ = (n $$-$$ m$$\Delta$$n)sin90$$^\circ$$ it is given that n = 1.6, $$\theta$$ = 30$$^\circ$$, $$\Delta$$n = 0.1. The above equation becomes
1.6sin30 = [1.6 $$-$$ m(0.1)]sin90
1.6 $$\times$$ $${1 \over 2}$$ = (1.6 $$-$$ 0.1m) $$\times$$ 1
0.8 = (1.6 $$-$$ 0.1m)
0.1m = 1.6 $$-$$ 0.8
0.1m = 0.8
m = 8
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