Graduate Why does the free end refraction occur after Brewster's angle in TM?

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The discussion centers on understanding why free end reflection occurs after the Brewster angle for transverse magnetic (TM) waves when transitioning from vacuum to a dense medium like glass. It highlights that at angles greater than the Brewster angle, the reflected wave experiences a 180-degree phase shift, leading to potential cancellation between incident and reflected waves. Participants suggest that the reflection coefficient can be modeled as a combination of free end and fixed end reflection terms, with the former dominating at larger angles. This dominance results in a positive reflection value, indicating free end reflection behavior. Overall, the interaction of phase shifts and wave cancellation is key to understanding this phenomenon.
jojoe
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Thanks for reading my question, and i’m really sorry about my poor english.

What i am wondering about is
why does the free end occur after the incidence angle overcome the Brewster angle in TM wave? (The ray incident from vacuum to glass(dense medium))

i tried to interpret this phenomenon with dipole oscillation of medium and the velocity difference of light in different medium(with more detail, vacuum to dense medium), but those trial didn’t make any effective conclusions.

Could someone help me about understanding this with macro or microscale aspect?
 
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Sorry Jojoe, can you try to find another word for free end as I cannot understand it.
At angles greater than the Brewster Angle, the reflected wave undergoes a 180 deg phase shift. At angles less than the Brewster Angle it does not. This phase shift at large angles of incidence can result in cancellation between incident and reflected waves.
 
tech99 said:
Sorry Jojoe, can you try to find another word for free end as I cannot understand it.
At angles greater than the Brewster Angle, the reflected wave undergoes a 180 deg phase shift. At angles less than the Brewster Angle it does not. This phase shift at large angles of incidence can result in cancellation between incident and reflected waves.
I think free end reflection might be proper representation, that the reflected light has the same phase with incident light. I thought about some idea that the reflection coefficient can be represented with the summation of free end reflection term(which has positive sign with cosine) and fixed end reflection term(which has negative sign of sine in sqrt). As the angle gets greater than Brewster angle, the free end reflection term is dominant, so the whole reflection terms gets positive value, which designates free end reflection of TM wave. Will it be related with the cancellation between the incident and reflected waves?
 

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Thank you, I understand now. I think you are correct. Cancellation occurs if the detector sees both the incident beam of light and the reflected beam of light for the case when angle of incidence is greater than the Brerwster Angle.
 
I do not have a good working knowledge of physics yet. I tried to piece this together but after researching this, I couldn’t figure out the correct laws of physics to combine to develop a formula to answer this question. Ex. 1 - A moving object impacts a static object at a constant velocity. Ex. 2 - A moving object impacts a static object at the same velocity but is accelerating at the moment of impact. Assuming the mass of the objects is the same and the velocity at the moment of impact...

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