Linear polarization in a glass prism

AI Thread Summary
The discussion centers on calculating the refractive index of a glass prism submerged in water, where unpolarized light reflects internally. The critical angle for total internal reflection is denoted as β, and the refractive index of water is given as 1.33. The participant initially struggles with the task of determining the refractive index without a specified wavelength but later realizes that the wavelength is not necessary for the calculation. They derive a formula involving the Brewster angle to find the refractive index, concluding that they may have solved the problem independently. The conversation highlights the relationship between the angles of incidence, reflection, and refraction in determining polarization.
Hatmpatn
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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 nv=1.33 and that the refractive index of the glass prism is greater than this(ng > nv).

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.

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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!
 
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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!
 
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.

c2QciUY.png


Unknown is ng.

I know that θ1+θ2=90°

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

The θ1 should be equal to our β.

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

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

Need some help here..
 
Last edited:
No need to help me here. Think I have solved it myself. I can post the solution tomorrow
 
Thread 'Variable mass system : water sprayed into a moving container'

Thread 'Variable mass system : water sprayed into a moving container'

Starting with the mass considerations #m(t)# is mass of water #M_{c}# mass of container and #M(t)# mass of total system $$M(t) = M_{C} + m(t)$$ $$\Rightarrow \frac{dM(t)}{dt} = \frac{dm(t)}{dt}$$ $$P_i = Mv + u \, dm$$ $$P_f = (M + dm)(v + dv)$$ $$\Delta P = M \, dv + (v - u) \, dm$$ $$F = \frac{dP}{dt} = M \frac{dv}{dt} + (v - u) \frac{dm}{dt}$$ $$F = u \frac{dm}{dt} = \rho A u^2$$ from conservation of momentum , the cannon recoils with the same force which it applies. $$\quad \frac{dm}{dt}...
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