Question

A beam of laser light is emitted 8cm below the surface of a mysterious liquid and strikes the surface (air above) at a point 7cm from the point directly above the laser. If total internal reflection occurs what is the refractive index of the mysterious liquid

Answer 1

here,

Applying Snell's Law to this situation,

sin i / sin r = n(r) / n(i),

When a beam strikes the surface at the critical angle for the liquid, the refracted beam travels along the surface, so the angle of the refracted ray is 90 degree,
hence sin r = 1.

From the given data, the incident ray makes an angle of arctan(7/8) with the normal, from which

i = arctan(7/8) = 41.2

sin i = 0.6585

Substituting in the equation

sin i / sin r = n(r)/n(i) :

0.6585 / 1 = 1 / n(i) . . . . . . (because the refractive index of air = 1)

Hence

n(i) = 1 / 0.6585 = 1.52

the refractive index of the liquid is 1.52