Air wedge - why is reflection from top of first slide ignored?

[Andrew Tom]

Homework Statement

Air wedge

Relevant Equations

$2t = n\lambda$ and $2t=(n+\frac{1}{2})\lambda$

My textbook derives the condition for bright and dark fringes on an air wedge by assuming that the reflected and refracted rays have a path difference of pi. Hence the conditions for bright and dark fringes end up being the opposite of what is expected.

However I did not really understand the derivation. The book says that the first ray will reflect from the BOTTOM of the top glass slide. So it is essentially reflected from glass, off air, hence there is no phase change. But I didn't understand why the book assumes the first ray is only reflected from the BOTTOM of the top glass slide. Isn't it also reflected from the top of the slide? This would mean it is reflected from air, off glass, hence it DOES undergo phase change of pi.

The second ray is refracted. It is then reflected from top of bottom glass slide. So it is reflected from air, off glass, hence it undergoes a phase change. Again, I am slightly confused why we can ignore the reflections of this second ray from the bottom of the bottom glass slide, and also its reflection from the bottom of the top glass slide, etc.

 

[Charles Link]

You are correct, there is also a reflection of light from the top surface of the top glass, and a reflection of light from the bottom surface of the bottom glass, but this light generally is fairly constant in intensity and doesn't result in generating any kind of interference patterns. If the glass is a very special type with extremely parallel faces, then it might be necessary to consider these other surfaces, but otherwise that is not the case.

Edit: There is another item that also needs to be considered to get any interference from these other two faces, and that is whether the source is sufficiently collimated. For the two inner surfaces, it is much easier to generate interference fringes, where the requirements aren't nearly as fussy.