Question

A layer of oil (n = 1.38) floats on an unknown liquid. A ray of light originates in the oil and passes into the unknown liquid. The angle of incidence is 65.0°, and the angle of refraction is 53.0°. What is the index of refraction of the unknown liquid?

Answer 1

Refractive index of unknown liquid = 1.56

Step-by-step explanation:

Using Snell's law as:

`n_i* {sin\theta_i}={n_r}*{sin\theta_r}`

Where,

`{\theta_i}` is the angle of incidence ( 65.0° )

`{\theta_r}` is the angle of refraction ( 53.0° )

`{n_r}` is the refractive index of the refraction medium (unknown liquid, n=?)

`{n_i}` is the refractive index of the incidence medium (oil, n=1.38)

Hence,

`1.38* {sin65.0^0}={n_r}*{sin53.0^0}`

Solving for
`{n_r}`,

Refractive index of unknown liquid = 1.56