Question

The absolute refractive index of water is root 2.what is the critical angle

Answer 1

Since the refractive index and critical angle are related as

n =1/ sin c

Where n is refractive index and c is critical angle.

So for n=1.414

Sin c =1/1.414 =0.707

c = 44.991°

Answer 2

Critical angle, $\theta_c=45^{\circ}$

Explanation:

Given that,

The absolute refractive index of water, $n=\sqrt{2}$

Let $\theta_c$ is the critical angle. We know that the relation between the critical angle and the refractive index is given by :

$n=\dfrac{1}{sin \theta_c}$

$sin \theta_c=\dfrac{1}{n}$

$\theta_c=sin^{-1}{\dfrac{1}{n}}$

$\theta_c=sin^{-1}{\dfrac{1}{\sqrt{2}}}$

$\theta_c=45^{\circ}$

So, the critical angle is 45 degrees.

Learn more :

Topic : Total internal reflection

https://brainly.in/question/1463662#