How Does Phase Accumulation Influence the Return Phase of a Wave?

[sapz]

Homework Statement

I hope this question is better suited to this forum...

(See picture)
We have 3 areas in which a wave can move.

The wave Y1 starts at area 1 and goes towards Border 1, some part of it is passed to Area 2.
That part goes towards Border 2, and some part of it is reflected back into Area 2.
That part moves towards Border 1, and some of it passes to Area 1.

I'm interested in that last part that returned to Area 1, which is Y4.
What is its Phase? It would seem that the wave accumulated phase when it was in Area 2, so should it be 2D * k2? (wave number times the distance in that area)?

If the original wave was Y_1(x,t) = Ae^{i(wt-k_1x)}, would Y4 be Y_4(x,t) = Be^{i(wt+k_1x+\phi)}, where \phi = 2Dk_2?
Or should it be \phi = -2Dk_2?

(A and B are some amplitudes we can relate through reflectivity and transmittance coefficients)

 

[haruspex]

The extra path length will create phase lag. Think of it this way, it has the same phase as the waves that were reflected from the first boundary at 2D/v in the past.
What about a possible 180 degree phase shift at the reflection?

 

[sapz]

The 180 degrees shift at reflection would be expressed by a minus sign in the reflection coefficient.
But what about the shift lag from the extra path? I am not sure I understand

 

[haruspex]

The 180 degrees shift at reflection would be expressed by a minus sign in the reflection coefficient.

Not necessarily. It depends on which medium has the higher index.

But what about the shift lag from the extra path? I am not sure I understand

A wave reflected at the second boundary has had to travel an extra 2D compared to a wave reflected at the first boundary. That takes time 2D/v. If the frequency at the source is ω, the phase at the source advances ω2D/v in that time. So the phase of a wave reflected at the second boundary will be ω2D/v behind a wave it meets that was reflected at the first boundary.