Question

How can you find critical angle of water? Explain your steps briefly.

Answer 1

We know the refractive index of water, $\mu =\frac{4}{3}=1.33$

And also the critical angle can be found by the formula,

$C={\sin }^{-1}\left(\frac{1}{\mu }\right)⟹C={\sin }^{-1}\left(\frac{1}{1.33}\right)⟹C={\sin }^{-1}\left(0.7518\right)⟹C={48.7}^{\circ }$

Hence the refractive index of water is ${48.7}^{\circ }$