A beam of light starts at a point 8.0 cm beneath the surface of a liquid and strikes the surface 7.0 cm from the point directly above the light source. Total internal reflection occurs for the beam. What can be said about the index of refraction of the liquid?

Explanation:

Given that,

The light starts at a point 8cm beneath the surface

It strikes the surface 7cm directly above the light

The angle of incidence from the given distances is

Using trigonometry

Tan I = opp / adj

tan I= (7.0 cm)/(8.0 cm) = 0.875

I = arctan(0.875)

I = 41.1859

According to Snell's law

n(water) sin I = n(air) sin R

Here R =90° for total reflection to occur

nair = 1

n(water) sin I = n(air) sinR

n(water) = (1)sin 90

n(water) = 1

n(water) = 1/sin I

n(water) =1/ sin(41.1859o)

= 1.52

The refractive index index of the liquid is 1.52

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