Question

A prism of refractive index \[\sqrt 2 \] has refracting angle of \[60^\circ \]. At what angle a ray must be incident on it so that it suffers a minimum deviation?
A. \[45^\circ \]
B. \[60^\circ \]
C. \[90^\circ \]
D. \[180^\circ \]

Answer 1

Hint:Use the equation that gives the relation between the angle of prism, angle of minimum deviation, angle of incidence and angle of emergence. Also use the formula for the refractive index of the material of the prism in terms of angle of minimum deviation. Use the condition for the minimum deviation of the incident light ray and solve.

Formulae used:
The relation between the angle of prism \[A\] and angle of minimum deviation \[{\delta _m}\] is
\[A + {\delta _m} = i + e\] …… (1)
Here, \[i\] is the angle of incidence and \[e\] is the angle of emergence.
The equation for the refractive index \[\mu \] of the material of prism is
\[\mu = \dfrac{{\sin \left( {\dfrac{{A + {\delta _m}}}{2}} \right)}}{{\sin \dfrac{A}{2}}}\] …… (2)
Here, \[A\] is the angle of prism and \[{\delta _m}\] is the angle of minimum deviation.

Complete step by step answer:
We have given that the refractive index of the material of the prism is \[\sqrt 2 \] and the angle of the prism is \[60^\circ \].
\[\mu = \sqrt 2 \]
\[\Rightarrow A = 60^\circ \]
We have asked to calculate the angle of incidence for which the deviation of the incident light is minimum.We know that in order to obtain minimum deviation of the incident light ray, the angle of incidence must be equal to the angle of emergence of the ray.
\[i = e\]
Substitute \[i\] for \[e\] in equation (1).
\[A + {\delta _m} = i + i\]
\[ \Rightarrow A + {\delta _m} = 2i\]
\[ \Rightarrow i = \dfrac{{A + {\delta _m}}}{2}\]
This is the expression for the angle of incidence of the light ray on the prism.

Let us now calculate the angle of incidence for which the deviation of the incident light ray is minimum. Substitute \[i\] for \[\dfrac{{A + {\delta _m}}}{2}\] in equation (2).
\[\mu = \dfrac{{\sin i}}{{\sin \dfrac{A}{2}}}\]
\[ \Rightarrow \sin i = \mu \sin \dfrac{A}{2}\]
\[ \Rightarrow i = {\sin ^{ - 1}}\left( {\mu \sin \dfrac{A}{2}} \right)\]
Substitute \[\sqrt 2 \] for \[\mu \] and \[60^\circ \] for \[A\] in the above equation.
\[ \Rightarrow i = {\sin ^{ - 1}}\left( {\sqrt 2 \sin \dfrac{{60^\circ }}{2}} \right)\]
\[ \Rightarrow i = {\sin ^{ - 1}}\left( {\sqrt 2 \sin 30^\circ } \right)\]
\[ \Rightarrow i = {\sin ^{ - 1}}\left( {\sqrt 2 \dfrac{1}{2}} \right)\]
\[ \Rightarrow i = {\sin ^{ - 1}}\left( {\dfrac{1}{{\sqrt 2 }}} \right)\]
\[ \therefore i = 45^\circ \]
Therefore, the angle of incidence to have minimum deviation of the light rays is \[45^\circ \].

Hence, the correct option is A.

Note: The students may get confused that the angle given in the question is the angle of refraction of the light ray as it is given as the refracting angle of the prism. But the students should keep in mind that the refracting angle of the prism is the angle of the prism. If one gets this wrong then one will not be able to solve the question.