Young's double slit experiment: Determine thickness of mica

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SUMMARY

The discussion focuses on determining the thickness of a mica sheet used in a double-slit experiment, specifically when the mica has a refractive index (n) of 1.582 and covers one slit. The central maximum is observed at a wavelength of 539 nm. To find the exact thickness, the optical path difference created by the mica must be calculated, considering that the optical path in air is treated as n=1 and the optical path in mica is n times the distance. The thickness of the mica can be derived from the condition that the optical path difference corresponds to an integral number of wavelengths.

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DriesBoon
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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? which is n times the distance of air? I understand i correctly
 

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