Angle of transmission past brewster's angle

In summary: When you have a ray (forget the polarization) coming to the interface between two media given the refractive indices and angle of incident, there is this equation you will always use to find the refracted angle."In summary, the author is trying to find the angle of transmission for a ray that has entered a medium with different refractive indices, but is not sure how to do it. They mention Fresnel's equations as a possible solution.
  • #1
lcr2139
62
1

Homework Statement


find TM p-parallel for (angle of incidence)=68 degrees, n(sub1) = 1.0, n(sub2) = 1.6[/B]

Homework Equations


how to find angle of transmission, theta(sub)i[/B]

The Attempt at a Solution


I know that when the angle of incidence is greater than the critical angle, all the light undergoes reflection. I have to use TM p-parallel equations. I know that using Brewster's angle, the angle of refraction is 90 degrees. How do I find the angle of transmission?[/B]
 
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  • #3
I am sorry but this does not answer my question. It only goes from lower than brewster's angle to brewster's angle. I am looking for an angle greater than brewster's angle. Can you help me?
 
  • #4
lcr2139 said:
find TM p-parallel
The question sounds a bit vague to me, are you asked to calculate the magnitude of the transmitted TM component? Then there is Fresnel equations as your machinery.
lcr2139 said:

Homework Equations


how to find angle of transmission, theta(sub)i[/B]
This part of the post should be allocated for the required equation to calculate the angle of transmission.
 
  • #5
I know I need to use fresnel's equations. My question is, how do I calculate the angle of transmission if it is greater than brewster's angle? Meaning, the angle of transmission is purely reflected.
 
  • #6
There are few things we need to correct before proceeding.
lcr2139 said:
I know that using Brewster's angle, the angle of refraction is 90 degrees.
No, it is not. When the angle of incidence is equal to the Brewster angle, it's the sum of the incident and refracted angles which forms 90 degree.
lcr2139 said:
I know that when the angle of incidence is greater than the critical angle, all the light undergoes reflection.
The total reflection only occurs for the case of internal reflection where the refractive index of the second medium is greater than the first medium. So, in your problem, total reflection will never be observed.
When you have a ray (forget the polarization) coming to the interface between two media given the refractive indices and angle of incident, there is this equation you will always use to find the refracted angle. Try to go back in your mind to your high school times to figure out this equation, if this does not work try googling "refraction".
 

FAQ: Angle of transmission past brewster's angle

1. What is the angle of transmission past Brewster's angle?

The angle of transmission past Brewster's angle is the angle at which light passes through a medium and is completely polarized. It is equal to the Brewster's angle plus 90 degrees.

2. How is Brewster's angle calculated?

Brewster's angle is calculated using the equation tan θ = n2/n1, where θ is the angle of incidence, n1 is the refractive index of the first medium, and n2 is the refractive index of the second medium.

3. What happens when light is incident at Brewster's angle?

When light is incident at Brewster's angle, it becomes linearly polarized and the reflected light is completely polarized perpendicular to the plane of incidence. This means the reflected light is fully polarized and does not have any components in the direction of the incident light.

4. Can light be transmitted at an angle past Brewster's angle?

Yes, light can be transmitted at an angle past Brewster's angle. This occurs when the angle of incidence is greater than Brewster's angle. In this case, the reflected light will still be partially polarized, but there will also be a component of light that is parallel to the plane of incidence.

5. Why is Brewster's angle important?

Brewster's angle is important because it allows us to control the polarization of light. By adjusting the angle of incidence, we can change the degree of polarization of the reflected light. This is useful in many applications, such as reducing glare on surfaces or in optical devices like polarizing filters.

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