Angle to the Normal and reflection/refraction HELP

[jkeyes2]

Homework Statement

Light incident from air strikes a diamond at 30°
a) At what angle to the normal a is the light reflected back into air?
b) At what angle to the normal a is the light refracted inside the diamond?
c) At what angle to the normal b is the light incident on the back side of the diamond?
d) At what angle to the normal b is the light reflected on the back side of the diamond?
e) at what angle to the normal b is the light refracted back into air on the back side of the diamond?

I am given that for the diamond, n= 2.4
The image given is attached.

Homework Equations

I thin my main question is what equation do I use? I think it is something like sinθ(n/something)

The Attempt at a Solution

I would really like to figure out the answers myself I think. I just need help getting started. Could you explain how to get started with this?

Any and all help is greatly appreciated. Thank you for your time! :-)

 

[anathema]

Hello,

To get started, we need to first understand the concept of Snell's Law. Snell's Law states that the ratio of the sine of the angle of incidence to the sine of the angle of refraction is equal to the ratio of the refractive indices of the two mediums. In this case, the two mediums are air and diamond.

So, for part a) we can use Snell's Law to find the angle of reflection. The angle of reflection is equal to the angle of incidence, so we can use the same value of 30° for both. The equation would be sin(30°) = (n_air/n_diamond) * sin(θ_reflection). Rearranging the equation, we get θ_reflection = sin^-1((n_air/n_diamond) * sin(30°)). Plugging in the values, we get θ_reflection = sin^-1((1/2.4) * sin(30°)) = 12.5°.

For part b), we can use the same equation but solve for the angle of refraction. The angle of refraction is given by θ_refraction = sin^-1((n_diamond/n_air) * sin(θ_incident)). Plugging in the values, we get θ_refraction = sin^-1((2.4/1) * sin(30°)) = 48.6°.

For part c), we need to consider the angle of incidence on the back side of the diamond. The angle of incidence on the back side is equal to the angle of refraction on the front side. So, the angle of incidence on the back side is also 48.6°.

For part d), we can use the same equation as part a) but with the angle of incidence on the back side. So, θ_reflection = sin^-1((n_air/n_diamond) * sin(48.6°)). Plugging in the values, we get θ_reflection = sin^-1((1/2.4) * sin(48.6°)) = 20.2°.

For part e), we can use the same equation as part b) but with the angle of incidence on the back side. So, θ_refraction = sin^-1((n_diamond/n_air) * sin(48.6°)). Plugging in the values, we get θ_ref