Question

what is the value of refractive index of the mediam if the critical angle of incident in a denser-rarer interface is equal to 45

Answer 1

Refractive index (n) and critical angle (c) are related as
=> n = 1 / Sinc

So,
n = 1 / Sin45
n = √2 = 1.414

∴ Refractive index of denser medium is 1.414

[Note: I assumed rarer medium as air]

Answer 2

Answer:

The value of refractive index is 1.414

Explanation:

sin 45 degree= 1/refractive index

so, refractive index = $\sqrt 2$

=1.414