I would like to solve the problem on the following image:

enter image description here

My question is: Why is the answer to (a) a minimum?

  1. When the light wave hits the top surface of the top glass, a wave will be reflected with phase change of pi.

  2. The non-reflected goes to the bottom surface of the top glass to be reflected again with no phase change.

  3. The non-reflected part still goes to the top of the down glass and is reflected with phase change pi.

  4. We don't care about other reflections as amplitude diminishes too much.

Now, We have three waves who are reflected towards our eye. 2 and 3 automatically cancel each other. There is only wave 1 left. So, this is not perfectly destructive. How can they call this a minimum?

  • $\begingroup$ Please help me out here. I am lost. $\endgroup$ – yolo123 Oct 9 '14 at 22:57
  • $\begingroup$ @ACuriousMind, any idea? $\endgroup$ – yolo123 Oct 9 '14 at 23:30
  • $\begingroup$ What's the phase change in the space between the two glass plates? $\endgroup$ – CuriousOne Oct 9 '14 at 23:33
  • $\begingroup$ pi. The two waves 2 and 3 cancel each other out. But wave 1 still stays. So how come it is a minima? $\endgroup$ – yolo123 Oct 9 '14 at 23:36
  • $\begingroup$ OK, I think I see your problem. In this case minimum doesn't mean total extinction. It just means that there is a visible variation in brightness. You can easily do this experiment yourself with some reasonably flat glass plates. As a kid I used to do it with glass from slide holders. It can probably be done nicely with microscope slides. What you will see in white light is a number of faint colored fringes. If you wanted near total extinction rather than slight variations in intensity, you would need mirrored and coated surfaces to get just one reflection. Does that help? $\endgroup$ – CuriousOne Oct 9 '14 at 23:42

For a start, and in context of question (a), you can simplify the situation by thinking it as two glass slabs parallel to each other with an air interface of neglictible thickness between them.

I'll consider only what happens in the first slab as by symmetry, what happens in the second one is the same.

Now, first: All the light you want to consider is the light that goes into the slabs. In effect, each time light meets an interface, part of it is transmitted and part of it is reflected. So think about the part of the light that is first transmitted through the top interface of the top glass (which is the great majority of it). Then, when it goes to the bottom*, part of it is transmitted and the rest is reflected back into the slab. This happens an infinite number of times (every time with a decreased intensity) on both interfaces so that in the end, the the light that comes back out from the slabs to your eyes is the majority of the light. Therefore, you can neglect your first bullet point. (I recommend you to look up the maths in your textbook, it's only a sum of an arithmetic progression if my memory doesn't betray me).

Second: Magic happens at the interface of both of the slabs. Go back to the * in the previous paragraph. At that time, two reflexions occur: one from the bottom of the top slab (with no phase shift as it is a glass/air interface) and one from the top of the bottom slab (with a phase shift as it is an air/glass interface). As we consider the air thickness to be neglictible, we obtain two waves with a Pi phase shift that destructively interfere. All you have to do is prove that the intensities of both of these are comparable (with the aforementioned sums) so that you can detect a minimum.

Hope that helps


enter image description here

This image (taken from the book Introduction to Optics by Pedrotti) illustrates what I mention in the first paragraph of my explanation: after multiple reflections and transmissions, the intensities of the rays add up. With this, you can figure out the sum I mentionned by yourself.

The r and t respectively stand for the reflection and transmission coefficients. Be careful with the t though as it is also used for the thickness of the slab (on the right side of the picture).

  • $\begingroup$ Mary, I don't understand why I can neglect "When the light wave hits the top surface of the top glass, a wave will be reflected with phase change of pi." It is still a wave! $\endgroup$ – yolo123 Oct 10 '14 at 0:15
  • $\begingroup$ "the majority of the light" : Does that not contradict what @CuriousOne said about light 4% reflection? $\endgroup$ – yolo123 Oct 10 '14 at 0:20
  • $\begingroup$ @yolo123 I don't have time to go into details right now but think in terms of intensities. This is in agreement with the 4% you mention: I said that the majority of the light is transmitted, isn't 96% the majority? $\endgroup$ – Mary Oct 10 '14 at 0:24
  • $\begingroup$ Oh... But what would happen if we were to replace the air inside the wedge with a material of, say, n=1.7? Would we still have 28 dark fringes? $\endgroup$ – yolo123 Oct 10 '14 at 0:27

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