Question

What is the critical angle for the interface between water and light flint? nflint= 1.58, nwater=1.33?

Answer 1

When light crosses the interface between a medium with higher refractive index and a medium with lower refractive index, there is a maximum value of the angle of incidence after which there is no refraction, but all the light is reflected, and this maximum value is called critical angle.

The critical angle is given by
$\theta_C = \arcsin ( \frac{n_2}{n_1} )$
where n1 is the refractive index of the first medium while n2 is the refractive index of the second medium. In our problem, n1=1.33 and n2=1.58, so the critical angle is
$\theta_C = \arcsin ( \frac{1.33}{1.58} )=57.3 ^{\circ}$