Question

# A beam of laser light is emitted 8cm below the surface of a mysterious liquid and...

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

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

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