Question

Follow the steps above to solve the problem: A diamond is underwater; A light ray enters one face of the diamond, then travels at an angle of 30 degrees with respect to the normal. What was the ray's angle of incidence on the diamond? The index of refraction of water is nwater=1.33, and the index of refraction of diamond is ndiamond=2.42.

Answer 1

65.48°

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 (? )

`{\theta_r}` is the angle of refraction ( 30.0° )

`{n_r}` is the refractive index of the refraction medium (diamond, n=2.42)

`{n_i}` is the refractive index of the incidence medium (water, n=1.33)

Hence,

`1.33* {sin\theta_i}={2.42}*{sin30^0}`

Angle of incidence =
`sin^(-1)0.9098` = 65.48°