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

  1. Apr 6, 2017 #1
    1. The problem statement, all variables and given/known data
    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

    2. Relevant equations
    ##E(x,t)=E_0 cos (kx-\omega t)##

    3. 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
     
  2. jcsd
  3. Apr 7, 2017 #2

    ehild

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    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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