FLP: Derivation of reflection coefficient

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SUMMARY

The discussion focuses on the derivation of the reflection coefficient as presented in Vol. I Ch.33 of Feynman's Lectures, specifically section 33-6. The confusion arises regarding the components of the electric field vector A in relation to the angle of incidence and reflection. It is clarified that the component of A perpendicular to B, represented as Acos(i+r), is essential for producing the electric field in the direction of B, while the component parallel to B does not contribute. Understanding these geometric relationships is crucial for grasping the underlying physics of light propagation.

PREREQUISITES
  • Familiarity with Feynman's Lectures on Physics, specifically Volume I, Chapter 33.
  • Understanding of vector components in physics, particularly in the context of light and polarization.
  • Knowledge of the principles of reflection and refraction of light.
  • Basic grasp of electric field vectors and their orientations.
NEXT STEPS
  • Study the derivation of the reflection coefficient in Feynman's Lectures, focusing on Chapter 33.
  • Explore the concept of polarization of light and its mathematical representation.
  • Learn about the geometric interpretation of vector components in optics.
  • Investigate the relationship between electric field vectors and light propagation direction.
USEFUL FOR

This discussion is beneficial for physics students, educators, and anyone interested in the principles of optics and light behavior, particularly in understanding reflection and polarization phenomena.

duffing
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In Vol. I Ch.33 of Feynman's Lectures (http://www.feynmanlectures.caltech.edu/I_33.html), 33-6, the reflection coefficient as a function of angle was derived.

I am confused about the part where it said the component of A perpendicular to B (Acos (i+r)) has the right polarisation to produce B. Geometrically I obtained Asin(i+r) instead but I am unsure about why the physics is so.

Similarly, it was also mentioned that the component of A normal to the dashed line (Acos(i-r)) is effective in producing the field of magnitude -1. I am confused as to why this component produces the field of magnitude -1.

Thanks.
 
duffing said:
I am confused about the part where it said the component of A perpendicular to B (Acos (i+r)) has the right polarisation to produce B. Geometrically I obtained Asin(i+r) instead but I am unsure about why the physics is so.

If you're asking about the physics rather than the specific formula, the electric field vector is always perpendicular to the direction of the ray. The component of A parallel to B cannot contribute to light in the B direction.
 

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