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

Tags:
1. Apr 6, 2017

### crick

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 $\pi/2$.

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. Apr 7, 2017

### ehild

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.