Fresnel Loss & Brewster's Angle

AI Thread Summary
Fresnel Loss occurs due to changes in refractive index, causing some light to be transmitted and some to be reflected, which can decrease laser efficiency. Brewster's Angle is the specific angle where no light is reflected, resulting in all exiting light being polarized. At Brewster's Angle, the reflected light is polarized, but the discussion raises a question about the polarization state of the light that passes through the medium. The reflectivity graph shows that parallel polarization reflectivity drops to zero at Brewster's Angle, while perpendicular polarization increases. Understanding these concepts is crucial for optimizing laser performance and minimizing losses.
says
Messages
585
Reaction score
12

Homework Statement


I'm trying to understand these two a bit better in terms of lasers.

Fresnel Loss - The loss that occurs when there is a change in refractive index. Some light will be transmitted, while other light will be reflected back into the medium. This is bad for lasers because it could reduce the efficiency of the laser.

Brewster’s Angle -- Is the angle where no light is reflected and all of the light that exits the medium and is polarized.

At Brewster's Angle, is all of the light exiting the medium polarized? or are there now two different polarizations exiting the medium? In the definitions I've seen for Brewster's Angle, they say the reflected light is now polarized and exits the medium, but they don't explicitly mention the 'other' light that originally went through the medium before Brewster's Angle was used to create the polarization.

2. Homework Equations

The Attempt at a Solution

 
Physics news on Phys.org
If you have a graph of reflectivity (R) vs. angle of incidence for each of parallel and perpendicular polarization , both go from low values at ## \theta=0 ## (normal incidence) to R=1.0 at ## \theta ##=90 degrees. The parallel polarization reflectivity goes down to R=0 at the Brewster angle before going to 1.0 at 90 degrees. The perpendicular polarization reflectivity monotonically increases. In both cases R+T=1.0 without any absorption losses.
 
Thread 'Collision of a bullet on a rod-string system: query'
In this question, I have a question. I am NOT trying to solve it, but it is just a conceptual question. Consider the point on the rod, which connects the string and the rod. My question: just before and after the collision, is ANGULAR momentum CONSERVED about this point? Lets call the point which connects the string and rod as P. Why am I asking this? : it is clear from the scenario that the point of concern, which connects the string and the rod, moves in a circular path due to the string...
Back
Top