Question

Incident ray comes to the glass sphere of refractive index \[\sqrt 2 \]. What will be the angle of deviation for the final ray coming out of the sphere.

Answer 1

Hint:
The above problem can be resolved using the concept and applications of Snell's Law. Snell's law is the basic law that involves the major concepts of the angles and the value of refractive indices. The mathematical formula for Snell's law is used in this problem, and then the value of the angle of refraction is calculated. Then, at last, using the formula for the angle of deviation of the light, the deviation angle is calculated.

Complete step by step solution
Given:
The angle of incidence is, \[i = 45^\circ \].
The refractive index of the glass is, \[{n_2} = \sqrt 2 \]
Then, apply the Snell’s law to find the angle of refraction as,
\[{n_1}\sin i = {n_2}\sin r\]
Here, \[{n_1}\] is the refractive index of air and its value is 1 and r is the angle of refraction.
Solve by substituting the values as,
\[
{n_1}\sin i = {n_2}\sin r\\
\left( 1 \right) \times \left( {\sin 45^\circ } \right) = \left( {\sqrt 2 } \right) \times \sin r\\
\sin r = \dfrac{1}{{\sqrt 2 }} \times \left( {\dfrac{1}{{\sqrt 2 }}} \right)\\
\sin r = \dfrac{1}{2}
\]
 Further solving the above result as,
\[
\sin r = \dfrac{1}{2}\\
\sin r = \sin 30^\circ \\
r = 30\;^\circ
\]
The total angle of deviation is given as,
\[
\delta = 2\left( {i - r} \right)\\
\delta = 2\left( {45^\circ - 30^\circ } \right)\\
\delta = 30\;^\circ
\]

Therefore, the angle of deviation for the final ray coming out of the sphere is \[30^\circ \].

Note:
To solve the given problem, we must go through the concept of the Snell's law. In the application of Snell's law, the standard value of the refractive index of air is utilized, and this needs to be remembered. Then the formula involves the angular representation of the incident ray and the reflected ray. These variables can be required to calculate or be given in the question to make the substitution, and the desired result could be determined easily.