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I am currently studying the textbook Modern optical engineering, fourth edition, by Warren Smith. Section 1.5 Interference and Diffraction says the following:

Now if the waves arrive at C in phase, they will reinforce; if they arrive one-half wavelength out of phase, they will cancel. In determining the phase relationship at C we must take into account the index of the material through which the light has traveled and also the phase change which occurs on reflection. This phase change occurs when light traveling through a low-index medium is reflected from the surface of a high-index medium; the phase is then abruptly changed by 180$^\circ$, or one-half wavelength. No phase change occurs when the indices are encountered in reverse order. Thus with the relative indices as indicated in Fig. 1.14, there is a phase change at C for the light following the A$^\prime$CD path, but no phase change at B for the light reflected from the lower surface. enter image description here As in the case of Young’s experiment described above, the difference between the optical paths ABC and A$^\prime$C determines the phase relationship. Since the index of refraction is inversely related to the velocity of light in a medium, it is apparent that the length of time a wave front takes to travel through a thickness $d$ of a material of index $n$ is given by $t =􏳲 nd/c$ (where $c \approx 3 􏳽\times 10^{10} \ \text{cm/s} = \text{velocity of light}$). The constant frequency of electromagnetic radiation is given by $c/􏳺\lambda$, so that the number of cycles which take place during the time $t =􏳲 nd/c$ is given by $(c/\lambda􏳺) \cdot (nd/c)$ or $nd/\lambda$􏳺. Thus, if the number of cycles is the same, or differs by an integral number of cycles, over the two paths of light traversed, the two beams of light will arrive at the same phase.
In Fig. 1.14, the number of cycles for the path $A^\prime C$ is given by $\dfrac{1}{2} + n_1 \dfrac{A^\prime C}{\lambda}$􏱘 (the one-half cycle is for the reflection phase change) and for the path $ABC$ by $n_2 \dfrac{ABC}{\lambda}$􏱘; if these numbers differ by an integer, the waves will reinforce; if they differ by an integer plus one-half, they will cancel.

As you can see, for the path $A^\prime C$, the equation for the number of cycles is given by $\dfrac{1}{2} + n_1 \dfrac{A^\prime C}{\lambda}$, where the thickness $d = A^\prime C$􏱘. What I don't understand is why it is said that the "thickness" is $A^\prime C$? What I mean is that, If we look at figure 1.14, we can see that that particular ray has constant medium until it reaches $C$ (that is, there is no change in medium for the ray from $A^\prime$ to $C$) so how does it make sense to treat this part of the ray as if there exists some medium with a "thickness"?

I suspect that the fact that we treat the "thickness" as starting at $A^\prime$ for the ray has something to do with the wave interpretation of light vs. the ray interpretation of light (and perhaps diffraction, as shown by Young's diffraction experiment?).

I would greatly appreciate it if people would please take the time to carefully explain this.

Related: If the number of cycles is the same, or differs by an integral number of cycles, then the two beams of light will arrive at the same phase

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Replace the word thickness with path length.

So the path length from $A^\prime$ to $C$ is $A^\prime C$ for a wave with wavelength $\dfrac{\lambda}{n_1}$ where $\lambda$ is the wavelength in a vacuum and $n_1$ is the refractive index.

This means that the number of waves in the length $A^\prime C$ is $\dfrac{A^\prime C}{\left(\dfrac{\lambda}{n_1}\right)}= n_1\dfrac{A^\prime C}{\lambda}$.
The fraction $\dfrac{1}{2}$ comes to compensate the $\pi$ phase change of the reflected wave as compared with the incident wave.

Note that $n_1\dfrac{A'C}{\lambda}$ is sometime written as $\dfrac{n_1\,A^\prime C}{\lambda}$, and $n_1\,A^\prime C$ is called the optical path.
This length in a medium of refractive index $n_1$ contains the same number of waves as a path length of $n_1\,A^\prime C$ in a vacuum.

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  • $\begingroup$ What do you mean that "the fraction $\dfrac{1}{2}$ comes to compensate the $\pi$ phase change of the reflected wave as compared to the incident way"? This sounds unusual. The reflected wave has its phase shifted by $\dfrac{\pi}{2}$. Apart from this, I think your answer is pretty good. $\endgroup$ – The Pointer Aug 7 at 17:41
  • $\begingroup$ @ThePointer Wikipedia - "Light waves change phase by 180° when they reflect from the surface of a medium with higher refractive index than that of the medium in which they are travelling. A light wave travelling in air that is reflected by a glass barrier will undergo a 180° phase change, while light travelling in glass will not undergo a phase change if it is reflected by a boundary with air." So in effect there is an extra half wavelength between the reflected rays at top and bottom of the block. $\endgroup$ – Farcher Aug 7 at 17:53
  • $\begingroup$ Oops, sorry, I meant $\dfrac{\lambda}{2}$ – not $\dfrac{\pi}{2}$. Anyway, I see what you meant now. $\endgroup$ – The Pointer Aug 7 at 17:58
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I don't see the exact part in the passage above (and the formatting is a bit odd), but yes it is relating the phase change over a distance/thickness in a particular medium.

The scenario is assuming that the waves arriving at both $A$ and $A'$ are in phase.

If phase change for $ABC$ is equal to the phase change for $A'C$, then the light will converge in phase. So the change is must be calculated over both paths and compared. We don't care about the path prior to $A$ and $A'$ because they are assumed to be equidistant and constant index.

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  • $\begingroup$ It's not clear to me how most of your answer is relevant to my question, except for the last part: "We don't care about the path prior to $A$ and $A'$ because they are assumed to be equidistant and constant index." Why does the path begin equidistant mean that we don't care about it? $\endgroup$ – The Pointer Aug 6 at 22:51
  • $\begingroup$ We're assuming some amount of coherence from the source. If so, the waves arriving at equidistant points from the source (while going through the same medium), will maintain spatial coherence because the travel time is identical. $\endgroup$ – BowlOfRed Aug 6 at 23:27
  • $\begingroup$ and how is spatial coherence relevant to this? $\endgroup$ – The Pointer Aug 6 at 23:31

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