Variable refractive index question 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 
.................………
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.
 A: The problem with your approach is that you’re not taking into account the fact that the refractive index is variable.
However, you can notice that the refractive index is increasing. Hence the angle of the light is also increasing. If the index was constant and equal to $\sqrt 2$, the angle of refraction would be equal to $45°$ as you noticed. As the index is less than $\sqrt 2$ for $0 <y<1$, the refraction angle is less than $45°$. However there is a deviation as the index is greater than $1$.
Hence the only possible answer is (C): $30°$.
This could be verified using an integral computation.
A: the answer will be D) no deviation .Snell's law is valid even if the refractive index is variable. It is always valid when normal to the junctions of different mediums are parallel to each other. so if a light has grazing incidence..it will emerge grazing in this case.
so at the second refraction angle of emergence will be 90 degrees, and angle of refraction will be 45. but your question is not clear what it is asking for . I think it is asking for emergence("angle of refraction when it comes out")
