Question

A light ray emanating from under water (index 1.333) makes an angle of 37.5° with the normal to the surface. What is the angle that the refracted ray makes with the normal as it emerges into air index (1.0003)?

Answer 1

here,

the refractive index of water , nw = 1.333

the refractive index of air , na = 1.0003

the angle of incidence , i = 37.5 degree

let the angle of refraction be r

using snell's law

sin(r)/sin(i) = nw/na

sin(r)/sin(37.5) = 1.33

solving for r

r = 54.06 degree

the angle of refraction is 54.06 degree