Question

The angle of deviation $\delta $ vs angle of incidence $i$ is plotted for a prism. Pick up the correct statements.

$\left( A \right)$ The angle of prism is ${60^ \circ }$
$\left( B \right)$ The refractive index of the prism is $n = \sqrt 3 $
$\left( C \right)$ For deviation to be ${65^ \circ }$ the angle of incidence ${i_1} = {55^ \circ }$
\[\left( D \right)\]The curve of $\delta $ vs. $i$ is a parabolic

Answer 1

Hint: When a light ray undergoes refraction, there will be some relation between the incident angle and emergent angle. When considering a prism, the minimum angle of deviation depends on the material of the prism. Apply the formula to find the minimum deviation. Then using minimum deviation, we can determine the refractive index of prism. Hence, we can determine ${i_1}$.

Formula used:
$\delta = i + e - A$
Where, $\delta $ is the angle of deviation, $i$ is the incident angle, $e$ is the emergent ray, $A$ is the angle of prism.

Complete step by step answer:
The light ray undergoes refraction when the light ray is incident on the surface of the prism. The ray of light bends towards the normal since the glass slab is the denser medium and it has a higher refractive index. That refracted light will act as the incident ray on the inner surface of the prism. The light ray will again undergo refraction and comes out the glass prism as an emergent ray.
Light bent through a smallest angle by an optical device is called angle of deviation. If the angle of incidence is equal to the emergent angle then the angle of deviation is minimum. When considering a prism, the minimum angle of deviation depends on the material of the prism.
$\delta = i + e - A$
$\delta $ is the angle of deviation, $i$ is the incident angle, $e$ is the emergent ray, $A$ is the angle of prism.
From the figure $\delta = {60^ \circ }$ and ${i_1} = {60^ \circ }$
Then minimum deviation $\delta = 2i - A$ or $A = 2i - \delta $
Then the angle of prism
$\Rightarrow A = \left( {2 \times 60} \right) - 60 = {60^ \circ }$
Refractive index is given by
$\Rightarrow n = \dfrac{{\sin \left( {\dfrac{{A + \delta }}{2}} \right)}}{{\sin \left( {\dfrac{A}{2}} \right)}}$
$\Rightarrow n = \dfrac{{\sin 60}}{{\sin 30}} = \sqrt 3 $
Then for angle of deviation ${65^ \circ }$, in the figure given the angle of emergence $e = {70^ \circ }$
$\Rightarrow {\delta _1} = {i_1} + e - A$
$\Rightarrow {i_1} = 65 - 70 + 60 = {55^ \circ }$

Hence all the options are correct.

Note: We should always remember that in case of glass slab, angle of emergence is always equal to incident angle. But when we consider prisms the angle of emergence is always equal to the incident angle only at minimum deviation. Light bent through a smallest angle by an optical device is called angle of deviation. When considering a prism, the minimum angle of deviation depends on the material of the prism.