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Question: A ray of light is incident at an angle of \[60^\circ \] on one face of a prism which has an apex ang...

A ray of light is incident at an angle of 6060^\circ on one face of a prism which has an apex angle 0f 3030^\circ . The ray emerging out of the prism makes an angle of 6060^\circ with the incident ray. The refractive index of that material of prism is-
A. 2\sqrt 2
B. 3\sqrt 3
C. 1.51.5
D. None of these

Explanation

Solution

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+δ=i1+i2A + \delta = {i_1} + {i_2} …… (1)
Here, AA is the angle of prism, δ\delta is the angle of deviation, i1{i_1} is the angle of incidence and i2{i_2} is the angle of emergence.
The equation for the angel of prism is
A=r1+r2A = {r_1} + {r_2} …… (2)
Here, AA is the angel of prism, r1{r_1} is the angle of emergence for the angle of incidence i1{i_1} on one face of prism and r2{r_2} is the angle of incidence on the second face of the prism.
The expression for the refractive index nn of the prism is
n=sini1sinr1n = \dfrac{{\sin {i_1}}}{{\sin {r_1}}} …… (3)
Here, i1{i_1} is the angle of incidence and r1{r_1} is the angle of emergence.
The angle of incidence on the one face of the prism is 6060^\circ .
i1=60{i_1} = 60^\circ
The angle of the prism and the angle of deviation is 3030^\circ .
A=30A = 30^\circ
δ=30\delta = 30^\circ
Determine the angel emergence i2{i_2} on the second face of the prism.
Rearrange equation (1) for i2{i_2}.
i2=A+δi1{i_2} = A + \delta - {i_1}
Substitute 3030^\circ for AA, 3030^\circ for δ\delta and 6060^\circ for i1{i_1} in the above equation.
i2=30+3060{i_2} = 30^\circ + 30^\circ - 60^\circ
i2=0\Rightarrow {i_2} = 0^\circ
Hence, the angle of emergence is 00^\circ .
Since the angle of emergence is 00^\circ , the angle of incidence for the second face of the prism will also be zero.
i2=0{i_2} = 0^\circ
Determine the angle of refraction for the angle of incidence r1{r_1}.
Substitute 00^\circ for r2{r_2} and 3030^\circ for AA in equation (2) and rearrange it for r1{r_1}.
30=r1+030^\circ = {r_1} + 0^\circ
r1=30\Rightarrow {r_1} = 30^\circ
Now calculate the refractive index of the prism.
Substitute 6060^\circ for i1{i_1} and 3030^\circ for r1{r_1} in equation (3).
n=sin60sin30n = \dfrac{{\sin 60^\circ }}{{\sin 30^\circ }}
n=3212\Rightarrow n = \dfrac{{\dfrac{{\sqrt 3 }}{2}}}{{\dfrac{1}{2}}}
n=3\Rightarrow n = \sqrt 3
Therefore, the refractive index of the prism is 3\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.