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

 

[HalfLostAndSearching]

The statement is asking for the refractive index of the glass at the current wavelength of the unpolarized light that is being reflected internally in the glass prism. This wavelength is not explicitly given, but it can be assumed to be within the visible range (400-700nm) since the statement mentions the critical angle of total reflection, which is typically observed in visible light.

To solve for the refractive index of the glass, we can use Snell's law, which 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 media. In this case, we have:

sin(i)/sin(r) = ng/nv

Where i is the angle of incidence (which is equal to the critical angle β in air), r is the angle of refraction (which is 90 degrees in this case since the light is being totally internally reflected), ng is the refractive index of the glass, and nv is the refractive index of water.

Since we know that the refractive index of water is 1.33, we can solve for the refractive index of the glass:

ng = sin(i)/sin(r) * nv

We can then plug in the value for the critical angle β in air, which can be found using the inverse sine function (sin^-1), and solve for the refractive index of the glass.

To answer part b, the result in a does not apply to light with arbitrary wavelengths because the critical angle and the resulting refractive index of the glass will vary depending on the wavelength of the light. This is due to the phenomenon of dispersion, where different wavelengths of light are refracted at different angles, resulting in different refractive indices.

For part c, diamonds have a higher refractive index than glass, so it is possible to observe total internal reflection in a diamond. However, the value of the refractive index will depend on the specific diamond and the wavelength of the light. A typical refractive index for diamond at a wavelength of 550nm is around 2.42. This can also be calculated using Snell's law, similar to part a.