Question on thin film interference

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SUMMARY

The discussion centers on calculating the number of interference fringes formed by a thin film of paper with a thickness of 15 x 10^-6 m, placed between two glass slides of length 7.25 cm, using light with a wavelength of 595 nm. The formula used is delta x = L (wavelength/2t), which yields a distance between dark fringes of 1.44 x 10^-3 m. Dividing the length of the glass slides by this distance results in 50 fringes, with a clarification that the total count is 51 due to the initial dark fringe. This confirms the correct interpretation of fringe counting in interference patterns.

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Homework Statement



A piece of paper of thickness 15 * 10^-6 m is used to create an air wedge between 2 glass slides that are 7.25 cm long. If the wavelength of the light being used is 595 nm, how many interference fringes are counted across the entire pattern?

Homework Equations



delta x = L ( wavelength/2t)

The Attempt at a Solution



delta x = (.0725)[(595 * 10^-9)/(2)(15 * 10^-6)]
= 1.44*10^-3 m

*now i know the distance between the dark fringes so i can divide the length of the glass slides to find the number of fringes:

.0725/1.44*10^-3
= 50 fringes

***is this correct?*** if it's wrong, a step-by-step solution would be greatly appreciated
 
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looks good to me. although i think it would be 51, because the pattern starts with a dark fringe, and then proceeds 50 dx's down the road before the last dark fringe. if you look at it like a fence, it takes 51 fence posts to hang 50 lengths of fence. the dx's are the lengths of fence and the dark fringes are the posts.

cheers
 

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