Phase shift of e.m. wave through a glass plate

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SUMMARY

The discussion focuses on the phase shift of an electromagnetic wave as it passes through a glass plate with a refractive index \( n \) and thickness \( \Delta x \). It establishes that if absorption is negligible, the wave received by an observer \( O \) experiences a phase difference of \( -\omega (n-1)\Delta x/c \) without altering the amplitude \( E_0 \). Additionally, for small phase shifts, the resultant wave can be expressed as the superposition of the original wave and a secondary wave with amplitude \( E_0 \omega (n-1) \Delta x/c \) and a phase shift of \( \pi/2 \).

PREREQUISITES
  • Understanding of electromagnetic wave propagation
  • Familiarity with the concept of refractive index
  • Knowledge of wave superposition principles
  • Basic proficiency in trigonometric identities and small angle approximations
NEXT STEPS
  • Study the derivation of phase shifts in optical media using Snell's Law
  • Learn about the application of the addition law for sine functions in wave interference
  • Explore the implications of small angle approximations in wave mechanics
  • Investigate the effects of varying refractive indices on wave propagation
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Students of physics, particularly those studying optics and wave mechanics, as well as educators seeking to clarify concepts related to phase shifts in electromagnetic waves.

crick
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Homework Statement


Consider a glass plate of refraction index n and thickness ## \Delta x ## placed between a point monochromatic source S and an observer O, as in picture.

(a) Prove that, if absorption from the plate is neglegible, then the effect on the wave received by O is the add of a phase difference ##-\omega (n-1)\Delta x/c## without changing the amplitude ##E_0## .

(b) If the phase shift is small, (##\Delta x \sim 0## or ##n \sim 1##) prove that the wave received by O can be seen as the superposition of the original wave of amplitude ##E_0## and a wave of amplitude ##E_0 \omega (n-1) \Delta x/c##, with a phase shift of ##\pi/2##.
Immagine.png


Homework Equations


##E(x,t)=E_0 cos (kx-\omega t)##

The Attempt at a Solution


For (a) I know that there is a path difference between a wave that propagates in air and a wave that propagrates in the plate which is ##n \Delta x## and from this derives a phase shift as in the text. Nevertheless I'm not really sure how can I prove it.

Anyway most of my doubts are about point (b). I find it difficult to understand where to start to prove this fact, any help is highly appreciated
 
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crick said:

Homework Statement


Consider a glass plate of refraction index n and thickness ## \Delta x ## placed between a point monochromatic source S and an observer O, as in picture.

(a) Prove that, if absorption from the plate is neglegible, then the effect on the wave received by O is the add of a phase difference ##-\omega (n-1)\Delta x/c## without changing the amplitude ##E_0## .

(b) If the phase shift is small, (##\Delta x \sim 0## or ##n \sim 1##) prove that the wave received by O can be seen as the superposition of the original wave of amplitude ##E_0## and a wave of amplitude ##E_0 \omega (n-1) \Delta x/c##, with a phase shift of ##\pi/2##.
View attachment 128401

Homework Equations


##E(x,t)=E_0 cos (kx-\omega t)##

The Attempt at a Solution


For (a) I know that there is a path difference between a wave that propagates in air and a wave that propagrates in the plate which is ##n \Delta x## and from this derives a phase shift as in the text. Nevertheless I'm not really sure how can I prove it.

Anyway most of my doubts are about point (b). I find it difficult to understand where to start to prove this fact, any help is highly appreciated
Assume the original wave ##E(x,t)=E_0 \sin (kx-\omega t)##. The glass slab causes a phase shift, so the new wave at O is ##E(x,t)=E_0 \sin (kx-\omega t + Φ)##. Apply the addition law for sin (α+β) taking α=kx-ωt and β=Φ=(n-1)Δxω/c. Use that the phase difference Φ is small, so cos(Φ)≈1 and sin(Φ)≈Φ.
 

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