Question

A monochromatic light is incident at a certain angle on an equilateral triangle prism and suffers minimum deviation. If the refractive index of the material of the prism is $\sqrt 3$, then the angle of incidence is?
A. ${30^0}$
B. ${45^0}$
C. ${90^0}$
D. ${60^0}$

Answer 1

In this question, we need to determine the angle of incidence of the monochromatic light on the equilateral triangle prism such that the refractive index of the prism is $\sqrt 3$. For this, we will use the properties of reflection along with the Snell’s Law.

Complete step by step answer:
The angle of incidence and the angle of emergence of the monochromatic light ray should be equal for the monochromatic light to have a minimum deviation when passing through an equilateral glass prism. Mathematically, $i = e$ where, ‘i' is the angle of incidence and ‘e’ be the angle of emergence.
Moreover, the angle of refraction in the equilateral prism for a monochromatic light ray should be half of the vertical angle of the prism. Mathematically, $r = \dfrac{A}{2}$ where, ‘r’ is the refraction angle and ‘A’ is the vertical angle of the prism.
According to the question, the prism is made up from the equilateral triangles so, the vertical angle of the prism is ${60^0}$. So, the angle of refraction is given by
$r = \dfrac{A}{2} \\\ = \dfrac{{{{60}^0}}}{2} \\\ = {30^0} \\\$
Now, following the Snell’s Law which that the product of the refractive index of the medium from where the incident light ray is coming and the sine of the angle of incidence will be equals to the refractive index of the medium where the refracted light travels and the sine of the angle of refraction. Mathematically, ${\mu _1} \times \sin i = {\mu _2} \times \sin r$ where, ${\mu _1}$ and ${\mu _2}$ are the refractive index of the medium from where the light is coming from and goes into respectively, ‘i' and ‘r’ are the angle of incidence and the angle of refraction.
Substituting the value of ${\mu _1} = 1;{\mu _2} = \sqrt 3$ and $r = {30^0}$ in the equation ${\mu _1} \times \sin i = {\mu _2} \times \sin r$ to determine the value of the angle of incidence.
${\mu _1} \times \sin i = {\mu _2} \times \sin r \\\ \Rightarrow 1 \times \sin i = \sqrt 3 \times \sin {30^0} \\\ \Rightarrow \sin i = \dfrac{{\sqrt 3 }}{2} \\\ \Rightarrow i = {\sin ^{ - 1}}\left( {\dfrac{{\sqrt 3 }}{2}} \right) \\\ \therefore i = {60^0} \\\$
Hence, the angle of incidence of the monochromatic light ray is ${60^0}$.

So, the correct answer is “Option D”.

Note:
Whenever a light rays travels from a denser medium (more refractive index) to the rarer medium (less refractive index) then, the light ray will bend away from the normal (line perpendicular to the plane joining both the mediums) while if the light rays travels from the rarer medium to the denser medium then, the light rays bend towards the normal.