# Variable refractive index question [closed]

A ray of light travelling in air is incident at grazing incidence on a slab with variable refractive index, $$n (y) = [k y^{3/2}+ 1]^{1/2}$$ where $$k = 1 m^{-3/2}$$ and follows path as shown in the figure. What is the angle of refraction when the ray comes out.

$$(A) 60^°$$
$$(B) 53^°$$
$$(C) 30^°$$
$$(D)$$ No deviation

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My approach:-

Let the angle of emergence br $$r$$
At origin i. e $$y=0$$ $$n_1=1$$ and when $$y=1$$ $$n_2=\sqrt2$$
Using Snell's law $$n_1 \sin90^° = n_2 sinr$$ On solving this I get $$r=45^°$$ but the sad part is that my answer doesn't match with any option.

## closed as off-topic by ACuriousMind♦Apr 26 at 16:12

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• The ray does not emerge into a medium of index $\sqrt{2}$. – nasu Apr 26 at 16:01
• So what index the ray emerge, I simply put y = 1 in given equation – Abhishek Kumar Apr 26 at 16:03
• Please note that homework-like questions and check-my-work questions are generally considered off-topic here. We intend our questions to be potentially useful to a broader set of users than just the one asking, and prefer conceptual questions over those just asking for a specific computation. – ACuriousMind Apr 26 at 16:12
• @Abhishek The ray emerges in air, according to your image. – nasu Apr 26 at 22:32

Answer is 4) No deviation . For a slab there is no deviation. $$n_1=n_2=1$$ of emergent ray from incident ray.