Thickness of Film:Constructive Interference @ 547nm, n=1.31

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The discussion focuses on calculating the smallest nonzero thickness of a soap film for constructive interference of reflected light at a wavelength of 547 nm and refractive index n=1.31. The formula used is 2t = (m + 1/2)(wavelength/nfilm), leading to a calculated thickness of 104 nm for m=0. Participants confirm that this calculation is correct. The thread also includes a request for verification of the solution. The conclusion affirms the accuracy of the result.
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Homework Statement


Light of wavelength 547nm is incident perpendicularly on a soap film (n=1.31) suspended in air. What is the smallest nonzero film thickness for which reflected light undergoes constructive interference?

Homework Equations


2t = (m + 1/2)(wavelength/nfilm)

The Attempt at a Solution


t = (0+1/2)(547nm/(2*1.31))
t = 104 nm

Is this correct?---------
Could someone please check this as well: https://www.physicsforums.com/showthread.php?t=310439
 
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It is correct.
 

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