Question

Light in vacuum is incident on the surface of a slab of transparent material. In the vacuum the beam makes an angle of 27.0° with the normal to the surface, while in the slab it makes an angle of 21.8° with the normal. What is the index of refraction of the transparent material?

Answer 1

Refractive index of slab = 1.22

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 ( 27.0° )

`{\theta_r}` is the angle of refraction ( 21.8° )

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

`{n_i}` is the refractive index of the incidence medium (vacuum, n=1)

Hence,

`1* {sin27.0^0}={n_r}*{sin21.8^0}`

Refractive index of slab = 1.22