Young's double slit experiment: Determine thickness of mica

AI Thread Summary
To determine the thickness of the mica sheet in Young's double slit experiment, the optical path difference must be calculated. The central maximum remains central for light with a wavelength of 539 nm, indicating that the optical path change due to the mica must equal an integral number of wavelengths. The optical path in air is simply the distance, while in mica, it is the thickness multiplied by the refractive index (n = 1.582). The discussion emphasizes the need to equate the optical path difference to an integer multiple of the wavelength to find the exact thickness of the mica sheet. Understanding these relationships is crucial for solving the problem effectively.
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Homework Statement


A sheet of mica ( approximate 6-7 µm) covers one slit
of a double-slit apparatus and has a n= 1.582. There is a central maximum of 539 nm. What is the exact thickness of the sheet of mica?

Homework Equations

The Attempt at a Solution


I think you can solve this with the number of wavelengths that go throug the sheet? But I'm a but stuck on the question.
 
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Welcome to the forum.

It seems like the question is implying that the central peak is still central for light of wavelength 539 nm. So the change in optical path when putting the mica on one side is an integral number of wavelengths. The optical path in air (treated as n=1, the n of air is something like 1.0003) is just the distance. The optical path in the mica with n>1 is n * the distance.

Is that enough of a hint?
 
So, I need to find the opitcal path in the mica? Wich is n times the distance of air? I understand i correctly
 
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