Question: - Assertion
The maximum refractive index of liquid for total internal reflection of ray passing from the prism should be $\sqrt 2 $
Reason
Here the value of critical angle is ${45^ \circ }$
A) Both Assertion and Reason are right and Reason is the correct explanation of Assertion.
B) Both Assertion and Reason are correct but reason is not the correct explanation of Assertion
C) Assertion is right but reason is wrong
D) Both Assertion and Reason are wrong.

Hint: Use the expression,
$\sin C = \dfrac{1}{\mu }$ (where, $C$is the critical angle and $\mu $is the refractive index)
Find the value of the critical angle by taking the maximum value of the refractive index.

Complete step by step answer:
The incident angle of light which travels in denser medium to make an angle of refraction of a less dense medium is called Critical Angle.
From definition,
$r = {90^ \circ }$
$i = C$(Where, C is the critical angle and i is the incident angle)
From Snell’s Law
$
  \dfrac{{sinC}}{{\sin r}} = \dfrac{{{\mu _1}}}{{{\mu _2}}} \\
  \dfrac{{sinC}}{{\sin {{90}^ \circ }}} = \dfrac{{{\mu _2}}}{{{\mu _1}}} \\
 $


For air, ${\mu _2} = 1$
$
  \sin C = \dfrac{1}{{{\mu _1}}} \\
   \Rightarrow C = {\sin ^{ - 1}}\left( {\dfrac{1}{{{\mu _1}}}} \right) \\
 $
When the angle of incidence is greater than critical angle during the travel of light, Total Internal Reflection occurs.
According to the Question,
${\mu _1} = \sqrt 2 $(where, $\mu $is refractive index)
$
  C = {\sin ^{ - 1}}\left( {\dfrac{1}{{\sqrt 2 }}} \right) \\
   \Rightarrow C = {\sin ^{ - 1}}\left( {\sin {{45}^ \circ }} \right) \\
   \Rightarrow C = {45^ \circ } \\
 $

Therefore, option (A) is correct because for the maximum value of refractive index the value of critical angle must be ${45^ \circ }$

Note: The refractive index of air is ${45^ \circ }$. In critical angle, the value of refraction becomes ${90^ \circ }$and incident angle becomes equal to critical angle. The value of the refractive index of air is always 1.