Soap Bubble interference-verify answer

  • Thread starter bcjochim07
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  • #1
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Homework Statement



A soap bubble is essentially a very thin film of water (n=1.33) surrounded by air. The colors thata you see in the soap bubbles are produced by interference.

Derive an expression for the wavelengths for which constructive interference casues a strong reflection from a soap bubble of thickness d.

Homework Equations





The Attempt at a Solution


2pi* m = 2pi*2d/(lambda/n) + intial phi = phase shift

In this case, when the light reflects off the bubble, the wave is shifted by pi, but when the waves go through the bubble reflect at the second water-air interface, the waves aren't shifted by pi, so the initial phase difference is equal to pi

2pi*m = 2pi*2dn/lambda + pi

pi(2m-1) = 4pi*dn/lambda

.5(m-.5) = dn/lambda

lambda= 2.66d/(m-.5) Is this right?
 

Answers and Replies

  • #2
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could someone please check my answer? I would really appreciate it. Thanks.
 
  • #3
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Please? I really would like to know.
 
  • #4
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ok firstly, are you only considering normal incidence?

secondly, I derived it and got:
[tex]d = (2m \pm 1)\frac{\lambda_s}{4}[/tex]
[tex]\lambda_s[/tex] is the wavelength in the soap.
Which is the same as yours if you don't divide everything by 2 in your equation.

Also, derivation for thin films is very nicely shown on page 393-395 in "OPTICS" by "E. HECHT"
 

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