When we consider the interference of light through a thin film of water or a glass slab, light gets reflected at both air-water and water-glass interface. Correspondingly, there is an extra phase difference of pi due to reflection at the denser medium for both rays . So the path difference for 1st ray should be $\lambda/2\mu_{air}$ and that for second wave should be $\lambda /2\mu_{water}$ . But in the derivation, this is not the case and they get cancelled out. Where is the flaw in my argument ?

My teacher said it has to do something with taking the path difference with respect to air for both rays but since the second ray travels in glass, shouldn't we just take it with respect to glass?

I am currently in high school.


1 Answer 1


You are right! You can see the following way:

Suppose the following setup :

Source: Optics by Ajoy ghatak

The path difference between the rays reflected from the upped surface and the lower surface is given by

$$\Delta x= 2\mu \lambda $$

and phase difference given by (with additional $\pi$ phase) : $$\Delta \phi=\frac{2\pi}{\lambda}\Delta x \pm \pi$$


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