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

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SUMMARY

The discussion centers on calculating the smallest nonzero thickness of a soap film that results in constructive interference for light of wavelength 547 nm, with a refractive index (n) of 1.31. The formula used is 2t = (m + 1/2)(wavelength/nfilm). The correct calculation yields a film thickness of 104 nm when m is set to 0. This confirms that the solution provided is accurate and adheres to the principles of optics.

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  • Understanding of constructive interference in optics
  • Familiarity with the refractive index concept
  • Knowledge of wavelength and its relation to film thickness
  • Ability to manipulate equations involving physical constants
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  • Study the principles of constructive and destructive interference in thin films
  • Learn about the effects of varying refractive indices on light behavior
  • Explore the application of the formula 2t = (m + 1/2)(wavelength/nfilm) in different contexts
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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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Anyone?
 
It is correct.
 

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