Question

A light source at the bottom of a 4.7-m-deep water pool sends a light ray up at an angle so that the ray strikes the surface 1.6 m from the point straight above the light source. What is the emerging ray's angle (in degrees) with the normal in air? °

Answer 1

According to snell's law: n₁ sin(i)=n₂ sin(r) where n₁ is refractive index of the medium in which incident ray is travelling ,n₂ is refractive index of the medium in which refracted ray is travelling , i is angle of incidence, r is angle of refraction.

Here,medium of incident ray is water whose refractive index is 4/3,i.e.,n₁=4/3 while medium of refraction is air whose refractive index is 1,i.e., n₂=1.

Also, tan(i)=(distance between point above the source and the point at which ray reaches interface)/depth=1.6/4.7

=>tan(i)=0.3404

=>i=tan^-1(0.3404)=18.8 degrees.

Using snell's law, 4/3*sin(18.8)=1*sin(r)

=>sin(r)=0.42969

=>r=sin^-1(0.42969)=25.25 degrees.

So, required angle=25.45 degrees.