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Wave Optics (thin films)

  • Thread starter lha08
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Homework Statement


A coating of film n=1.37 on glass slabs (n=1.6) is 8.45X10^-5 cm thick.If white light is incident normally, which visible wavelengths are missing in the reflected light? White light containing wavelengths from 400 to 700 nm.


Homework Equations





The Attempt at a Solution


I know how to solve for the answer but how am i supposed to know whether it is constructive and destructive? Also, what does the phase changes mean (e.g. pi and 0) and does it have any importance when i solve for the answer?? Thanks!!
 

Answers and Replies

  • #2
Doc Al
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I know how to solve for the answer but how am i supposed to know whether it is constructive and destructive?
Note that the problem asks about wavelengths that are missing in the reflected light.
Also, what does the phase changes mean (e.g. pi and 0) and does it have any importance when i solve for the answer??
When light reflects off of a surface with a higher index of refraction, there's a phase change of pi. So, in this case, both reflections get the same phase change.
 
  • #3
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Note that the problem asks about wavelengths that are missing in the reflected light.

When light reflects off of a surface with a higher index of refraction, there's a phase change of pi. So, in this case, both reflections get the same phase change.
So, does that mean that it depends on whether if they use "missing" or "enhanced" that allow us to determine if it's constructive or destructive? missing=destructive enhanced=constructive?

I'm not really sure but if i was to graph this, would i ultimately have 2 sinusoidal functions where the first function would have a phase change of pi and the second one with pi as well? so it would just look like a single function that begins at pi?
Thanks a lot.
 
  • #4
Doc Al
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So, does that mean that it depends on whether if they use "missing" or "enhanced" that allow us to determine if it's constructive or destructive? missing=destructive enhanced=constructive?
That's right.
I'm not really sure but if i was to graph this, would i ultimately have 2 sinusoidal functions where the first function would have a phase change of pi and the second one with pi as well? so it would just look like a single function that begins at pi?
What matters is the phase difference between the reflected waves. That difference has two sources: (1) phase change upon reflection, and (2) path length differences. Since (1) is the same for both waves, all you need to worry about is (2).
 
  • #5
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What matters is the phase difference between the reflected waves. That difference has two sources: (1) phase change upon reflection, and (2) path length differences. Since (1) is the same for both waves, all you need to worry about is (2).
Is the phase change and the path difference the same thing except that phase change is in radians and path difference is in meters? Other than that, i'm not really certain what the difference between the two really is..
 
  • #6
Doc Al
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Is the phase change and the path difference the same thing except that phase change is in radians and path difference is in meters? Other than that, i'm not really certain what the difference between the two really is..
The phase change upon reflection has nothing to do with path length difference, but both factors together determine the phase difference (measured in radians) between the reflected waves. The path length difference is due to one of the reflected waves traveling a longer distance since it goes back and forth through the thin film; that extra distance translates into a phase difference.
 

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