Calculating Degree of Polarization at Air-Glass Interface

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SUMMARY

The discussion focuses on calculating the degree of polarization of light at an air-glass interface, specifically at an incidence angle of 40 degrees with glass refractive index n = 1.5. The degree of polarization is defined by the formula I(polarized) / [I(polarized) + I(unpolarized)]. Participants emphasize the importance of using Fresnel relations to determine the transmission and reflection coefficients for both polarizations of light. Numerical evaluations for transmitted and reflected light polarization are also discussed, particularly at Brewster's angle.

PREREQUISITES
  • Understanding of electromagnetic fields associated with light
  • Familiarity with Fresnel relations for reflection and transmission
  • Knowledge of Brewster's angle and its significance in polarization
  • Basic concepts of intensity ratios in polarized and unpolarized light
NEXT STEPS
  • Study the derivation and application of Fresnel equations for light at interfaces
  • Learn how to calculate Brewster's angle for different materials
  • Explore numerical methods for evaluating intensity ratios in polarization
  • Investigate the effects of varying incidence angles on light polarization
USEFUL FOR

Students and professionals in optics, physics, and engineering fields, particularly those studying light behavior at material interfaces and polarization effects.

TimNguyen
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A beam of natural light is incident (from the air) on an air-glass interface at 40 degrees. Take n = 1.5 for the glass.

a)Obtain an expression for the degree of polarization of the transmitted light and compute it numerically.

b)Obtain an expression for the degree of polarization of the reflected light and compute it numerically.

c)Obtain an expression for the degree of polarization of the transmitted light when the angle of incidence is Brewster's angle and evaluate it numerically.

The only formula that I'm given is that the degree of polarization is equal to I(polarized) / [I(polarized) + I(unpolarized)].

How do I know which light is polarized and obtain the intensity for each polarized/unpolarized light?
 
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Could anyone help out?
 
It might help if you mentioned what kind of class you're taking and what sort of other things you've been studying.

I'm going to guess that you're taking a class where you're discussing the electric and magnetic fields associated with light.

The problem amounts to determining the boundary conditions between E&M in air and E&M in glass. The solution is that the tangential E&M fields are continuous and that this is enough to define the changes in angles.

You will have to work it out twice, once with light polarized one way and once with light polarized the other. That is, one way the E field is tangential to the surface, the other way the B field is tangential.

After you get the transmission and reflection coefficients for the two ways light can arrive, you will see that there is a difference in how much they reflect and transmit. That makes the light that is reflected polarized even if the light coming in wasn't. That's what you're calculating.

Good luck, it's a tough assignment.

Carl
 
TimNguyen said:
The only formula that I'm given is that the degree of polarization is equal to I(polarized) / [I(polarized) + I(unpolarized)].

How do I know which light is polarized and obtain the intensity for each polarized/unpolarized light?


You must have had something that's called "reflexion and transmission at planar interfaces", no ?
Look for something called "Fresnel relations" which give you the relative intensities of reflected and transmitted light incident under an angle for the two polarizations (perpendicular and parallel to the reflecting surface).

As CarlB suggests, consider the incident light as a 50-50 mix of both, and then see what fraction survives after reflexion or transmission and how much it deviates from this 50-50 ratio in both cases.

cheers,
Patrick.
 
The book claims the answer is that all the magnitudes are the same because "the gravitational force on the penguin is the same". I'm having trouble understanding this. I thought the buoyant force was equal to the weight of the fluid displaced. Weight depends on mass which depends on density. Therefore, due to the differing densities the buoyant force will be different in each case? Is this incorrect?

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