Question

A ray of light undergoes deviation of \[30^{\circ}\] when incident on an equilateral prism of refractive index $\sqrt 2 $​. The angle made by the ray inside the prism with the base of the prism is:

Answer 1

Hint: Here we use Snell’s law and expression for finding the deviation angle of a prism to solve this question. The law states that the ratio of the sin angle of incidence to the sine angle of refraction gives the refractive index of the medium. The Refractive index is a dimensionless quantity that gives information about the speed of light traveling through any material.

Formula used:
Refractive index of a prism:
$\mu = \dfrac{{\sin (\dfrac{{A + {\delta _m}}}{2})}}{{\sin \dfrac{A}{2}}}$
Where,
A- The angle of the prism
\[\delta_m\]- The minimum deviation of the prism

Complete answer:
An equilateral triangular prism is defined as a prism that has two parallel and congruent equilateral triangular prisms with an angle of 60 degrees and three rectangular faces perpendicular to the triangular faces. We are given that prism is an equilateral prism. So, the angle of the prism A will be 60 degrees.

Now we need to find minimum deviation. We know that minimum deviation is denoted by ${\delta _m}$ and it can be found by the relation between refractive index and minimum derivation for a prism,
$\mu = \dfrac{{\sin (\dfrac{{A + {\delta _m}}}{2})}}{{\sin \dfrac{A}{2}}}$

By substituting the values $A = 60 \circ $ and $\mu = \sqrt 2 $ we get,
$\sqrt 2 = \dfrac{{\sin (\dfrac{{60 + {\delta _m}}}{2})}}{{\sin \dfrac{{60}}{2}}}$
$ \Rightarrow \sqrt 2 \times \dfrac{1}{2} = \sin (\dfrac{{60 + {\delta _m}}}{2})$
$ \Rightarrow \dfrac{1}{{\sqrt 2 }} = \sin (\dfrac{{60 + {\delta _m}}}{2})$
$ \Rightarrow {\sin ^{ - 1}}\dfrac{1}{{\sqrt 2 }} = \dfrac{{60 + {\delta _m}}}{2}$Solving we get, ${\delta _m} = 30 \circ $

Here we can see that minimum deviation angle is equal to deviation angle. So we can say that rays travels parallel to base of equilateral prism.(From minimum deviation condition)

So, the angle made by the ray and base of the prism is zero.

Note: Do not confuse reflection with refraction. This is because refraction tells us about the bending of the path of light while entering into a medium while reflection determines the striking of light to an object and the idea about the path of light moving away from the reflecting object. Reflection phenomena happen within the same medium.