Linear polarization in a glass prism

[Hatmpatn]

Homework Statement

Unpolarized light is reflected internally in the point P in a glass prism. When the prism is located in air, β is the critical angle of total reflection.

I am going to calculate for the following problems:

a) If the prism is submerged into water the reflected light becomes completely linear polarized. Determine the refractive index of the glass at the current wavelength.
Assume that the water has the refractive index n_v=1.33 and that the refractive index of the glass prism is greater than this(n_g > n_v).

b) Does the result in a apply to light with arbitrary wavelengths? Why?

c) Assume that you only have access to diamonds instead of glass prisms. Can you make a similar experiment as in task a? Find out a typical refractive index for diamond when the light has the wavelength λ = 550nm.

The Attempt at a Solution

No attempt has yet been made, since I don't understand the sentence "Determine the refractive index of the glass at the current wavelength." in task a. This because I don't see any given wavelength. If anyone care to explain this I would be grateful!

 

[Hatmpatn]

Apparently, you don't need to know the wavelength of task a and therefore it is not given!
I'll try to come up with an attempt at the solution later today!

 

[Hatmpatn]

Alright, here is my first attempt at a solution:

Because the ng > nv, the outgoing beam of light will be reflected away from the normal of the plane(blue arrow).

Some of the light will leave the prism(refracted light) and some will be reflected(reflected light). If the angle of the refracted and reflected light is equal to 90°, then the light that is reflected will be completely polarized.

I need to calculate the Brewster angle. The angle where the incident light results in the 90° angle between the refracted and reflected light.

I have made notations in the image below, where the green arrow is the reflected light, the red is the refracted light and the blue is the normal to the plane of the prism.

Unknown is n_g.

I know that θ₁+θ₂=90°

Using the snells law: n_g*sin(θ₁)=1.33*sin(θ₂)
=> n_g*sin(θ₁)=1.33*sin(90°-θ₁)
=> n_g*sin(θ₁)=1.33*cos(θ₁)

The θ₁ should be equal to our β.

So the solution for the refractive index n_g, should be n_g*sin(β)=1.33*cos(β)
<=> n_g=1.33*cos(β)/sin(β)

But I don't think that our β is correct, sine that is the β when the our n_v is air...

Need some help here..

 

[Hatmpatn]

No need to help me here. Think I have solved it myself. I can post the solution tomorrow