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Question: A ray of light is incident normally on the surface of an equilateral prism made up of material with ...

A ray of light is incident normally on the surface of an equilateral prism made up of material with refractive index 1.5. The angle of deviation is
(A) 3030^\circ
(B) 4545^\circ
(C) 6060^\circ
(D) 7575^\circ

Explanation

Solution

If the angle of incidence is greater than the critical angle of the medium then the light ray is totally reflected back. First, we will use Snell’s law to find the angle of refraction and observe the behavior of the ray at the second face.
Angle of deviation δ=i+eA\delta = i + e - A

Complete step by step solution:
We can observe the angle of incidence i=0i = 0^\circ since it is given that the ray of light is normally incident on the surface of the prism. Also, the angle of the prism A=60A = 60^\circ .
We know that from Snell’s law sinr1=sinin\sin {r_1} = \dfrac{{\sin i}}{n} .
sinr1=sin0n=0\Rightarrow \sin {r_1} = \dfrac{{\sin 0^\circ }}{n} = 0
r1=0\Rightarrow {r_1} = 0^\circ
Now, we also know that for a prism A=r1+r2A = {r_1} + {r_2} .
r2=Ar1=600\Rightarrow {r_2} = A - {r_1} = 60^\circ - 0^\circ
r2=60\Rightarrow {r_2} = 60^\circ
We also know that if cc is the critical angle of the material of the prism, then sinc=1n\sin c = \dfrac{1}{n} .
sinc=11.5=1(32)=23\Rightarrow \sin c = \dfrac{1}{{1.5}} = \dfrac{1}{{\left( {\dfrac{3}{2}} \right)}} = \dfrac{2}{3}
c=sin1(23)\Rightarrow c = {\sin ^{ - 1}}\left( {\dfrac{2}{3}} \right)
c=41.81\Rightarrow c = 41.81^\circ
So, the critical angle of the material of the prism is 41.8141.81^\circ .
Here, the angle of incidence at the second face of the prism r2=60{r_2} = 60^\circ is greater than the critical angle of the prism. So, total internal reflection happens at the second face of the prism. That is, the light ray is totally internally reflected back into the prism from the second face. The angle of reflection will be 6060^\circ as r2=60{r_2} = 60^\circ . This ray will be incident normally on the base face of the prism. Hence, it will emerge normally perpendicular to the prism. That is, it will emerge with the angle of emergence e=0e = 0^\circ .
Now we know that δ=i+eA\delta = i + e - A .
δ=0060.\Rightarrow \delta = 0 - 0 - 60.
So, the angle of deviation will be 6060^\circ. Hence, the correct option is (C).

Note:
The angle of deviation steadily decreases as the angle of incidence is increased. But after attaining a certain minimum value, it starts increasing with increase in the angle of incidence. This minimum value is called the angle of minimum deviation and is denoted by DD.