Soap Bubble interference-verify answer

  • Thread starter Thread starter bcjochim07
  • Start date Start date
  • Tags Tags
    Bubble Soap
Click For Summary

Homework Help Overview

The problem involves deriving an expression for the wavelengths that lead to constructive interference in a soap bubble, which is modeled as a thin film of water surrounded by air. The context is related to optics and wave interference phenomena.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • The original poster attempts to derive a formula for the wavelengths using phase shifts and interference conditions. Some participants question whether the derivation considers normal incidence and suggest alternative expressions for the relationship between thickness and wavelength.

Discussion Status

The discussion is ongoing, with participants providing feedback on the original poster's derivation. Some have offered alternative formulations and references for further reading, indicating a productive exchange of ideas without reaching a consensus.

Contextual Notes

There is a mention of specific conditions such as normal incidence and the use of different wavelengths in the soap, which may affect the derivation. The original poster's request for verification indicates a need for clarity on the correctness of their approach.

bcjochim07
Messages
366
Reaction score
0

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?
 
Physics news on Phys.org
could someone please check my answer? I would really appreciate it. Thanks.
 
Please? I really would like to know.
 
ok firstly, are you only considering normal incidence?

secondly, I derived it and got:
d = (2m \pm 1)\frac{\lambda_s}{4}
\lambda_s 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"
 

Similar threads

Replies
3
Views
3K
  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 5 ·
Replies
5
Views
9K
  • · Replies 4 ·
Replies
4
Views
7K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 1 ·
Replies
1
Views
3K
Replies
2
Views
4K
Replies
2
Views
4K
  • · Replies 1 ·
Replies
1
Views
1K
Replies
9
Views
6K