A ray of light traveling in air strikes the surface of a liquid. if the angle of incidence is 31.7◦ and the angle of refraction is 21.3◦, find the critical angle for light traveling from the liquid back into the air. answer in units of ◦.

Media 1 and 2 are air and liquid. By Snell's law;

n1/n2 = Sin ∅2/Sin ∅1

Then,

n2 = (n1 * Sin Ф1) / Sin ∅2 = (1*Sin 31.7) / Sin 21.3 = 1.4466.

When the light travels in the opposite direction and at critical angle, media 1 and 2 are liquid and air respectively while ∅2 = 90°

Therefore,

n1/n2 = Sin 90 / Sin ∅c = > ∅c = Sin ^-1[n2*Sin 90]/n1 = Sin ^-1[1*Sin 90]/1.4466 = 43.73°

The critical angle (∅c) is 43.73°.