Question

A ray of light is incident at an angle of \[60^\circ \] on one face of a prism which has an apex angle 0f \[30^\circ \]. The ray emerging out of the prism makes an angle of \[60^\circ \] with the incident ray. The refractive index of that material of prism is-
A. \[\sqrt 2 \]
B. \[\sqrt 3 \]
C. \[1.5\]
D. None of these

Answer 1

Hint: Use the formula for the relation between the angle of prism, angle of deviation, angle of incidence and angle of emergence for the prism. Also use the relation between the angle of prism, angle of incidence and angle of emergence. Use the formula for the refractive index of the prism.

Complete Step by step answer: The relation between the angle of emergence, angle of deviation and angle of prism is given by
\[A + \delta = {i_1} + {i_2}\] …… (1)
Here, \[A\] is the angle of prism, \[\delta \] is the angle of deviation, \[{i_1}\] is the angle of incidence and \[{i_2}\] is the angle of emergence.
The equation for the angel of prism is
\[A = {r_1} + {r_2}\] …… (2)
Here, \[A\] is the angel of prism, \[{r_1}\] is the angle of emergence for the angle of incidence \[{i_1}\] on one face of prism and \[{r_2}\] is the angle of incidence on the second face of the prism.
The expression for the refractive index \[n\] of the prism is
\[n = \dfrac{{\sin {i_1}}}{{\sin {r_1}}}\] …… (3)
Here, \[{i_1}\] is the angle of incidence and \[{r_1}\] is the angle of emergence.
The angle of incidence on the one face of the prism is \[60^\circ \].
\[{i_1} = 60^\circ \]
The angle of the prism and the angle of deviation is \[30^\circ \].
\[A = 30^\circ \]
\[\delta = 30^\circ \]
Determine the angel emergence \[{i_2}\] on the second face of the prism.
Rearrange equation (1) for \[{i_2}\].
\[{i_2} = A + \delta - {i_1}\]
Substitute \[30^\circ \] for \[A\], \[30^\circ \] for \[\delta \] and \[60^\circ \] for \[{i_1}\] in the above equation.
\[{i_2} = 30^\circ + 30^\circ - 60^\circ \]
\[ \Rightarrow {i_2} = 0^\circ \]
Hence, the angle of emergence is \[0^\circ \].
Since the angle of emergence is \[0^\circ \], the angle of incidence for the second face of the prism will also be zero.
\[{i_2} = 0^\circ \]
Determine the angle of refraction for the angle of incidence \[{r_1}\].
Substitute \[0^\circ \] for \[{r_2}\] and \[30^\circ \] for \[A\] in equation (2) and rearrange it for \[{r_1}\].
\[30^\circ = {r_1} + 0^\circ \]
\[ \Rightarrow {r_1} = 30^\circ \]
Now calculate the refractive index of the prism.
Substitute \[60^\circ \] for \[{i_1}\] and \[30^\circ \] for \[{r_1}\] in equation (3).
\[n = \dfrac{{\sin 60^\circ }}{{\sin 30^\circ }}\]
\[ \Rightarrow n = \dfrac{{\dfrac{{\sqrt 3 }}{2}}}{{\dfrac{1}{2}}}\]
\[ \Rightarrow n = \sqrt 3 \]
Therefore, the refractive index of the prism is \[\sqrt 3 \].

Hence, the correct option is B.

Note: It should also be noted that since the angle of incidence for the second face of prism is zero, the angle of emergence of the second face is also zero. While finding out the refractive index students make mistakes while putting the correct value of the terms at exact position.