Solving Thin-Film Problem with Wavelengths 690 and 575nm

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The discussion focuses on solving a thin-film interference problem involving wavelengths of 690nm and 575nm, with a film index of refraction at 1.25 and water at 1.33. The key insight is that for constructive interference, the thickness of the film can be determined using the formula for optical path difference. Specifically, the thickness can be calculated using the relationship between wavelength and the film's refractive index, leading to two equations for the two wavelengths provided.

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Problem states: Wavelengths of 690 and 575nm are observed when white light is directed on a film floating in water. The index of the film is 1.25, and the index of water is 1.33.

What is the thichkness?

I know the phase is shifted by pi., but other than that I am not sure where to go with this problem!

Can I get a nudge in the right direction?

Thanks

--k
 
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I suppose I am only searching for the equation used to determine the answer.

I think that for thin films which have an index of refraction higher than the water, the thickness is wavelength over 4... but, for the situation of having an index lower than that of the water, I presume that some light makes it to/through the water...

Any help would be greatly appreciated!
 
Some light does make it through, even if the index is higher. But they're not asking about the amount of light transmitted or reflected - what they are giving you are the wavelengths where the reflection is a maximum. What is the name of chapter you working on now - interference? You have constructive interference at both wavelengths, which implies something about the thickness of the film - so you should be able to set up 2 equations relating wavelength, thickness and optical path difference between waves reflecting off the front and back surfaces of the film.
 

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