Question

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. Finally, apply Snell's law and find the ray's angle of incidence θ1 on the diamond.

Answer 1

According to snells law 

n1 sin theta1 = n2 sin theta2
n1  = 1.333 (water)
n2  = 2.42 (diamond)
 it is given that theta =30 degrees so 
by putting the values we have 

1.333 sin theta = 2.42 sin 30 

sin theta = (2.42/1.333) *0.5 =65.2 degree
so our conclusion is 

the ray's angle of incidence θ1 on the diamond = 65.2 degree.
hope this helps