How do I solve this Thick Film, Double Slit problem?

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Homework Help Overview

The discussion revolves around a thick film double slit problem in optics, specifically focusing on the phase change of light as it travels through a medium with a given refractive index.

Discussion Character

  • Exploratory, Assumption checking

Approaches and Questions Raised

  • Participants discuss the relationship between the thickness of the film and the phase change of light, with some questioning how to relate thickness to phase change in order to determine the minimum possible thickness.

Discussion Status

Some participants have provided insights into the phase change formula, while others have expressed confusion about the relevance of thickness to phase change. One participant reports having solved part of the problem, indicating some progress in the discussion.

Contextual Notes

There is a mention of a figure attached to the problem statement, which may contain additional context or constraints relevant to the discussion.

jegues
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Homework Statement



See figure attached for problem statement as well as my attempt.

Homework Equations





The Attempt at a Solution



I don't know where to go from here. Can someone help me get started?
 

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When light traverses x distance in a medium of refractive index n its phase changes by Edit: 2pi*n*x/lambda (lambda is the wavelength in vacuum).

ehild
 
Last edited:
ehild said:
When light traverses x distance in a medium of refractive index n its phase changes by 2pi*n/lambda (lambda is the wavelength in vacuum).

ehild

So the thickness doesn't relate to its phase change at all? How am I suppose to solve the minimum possible thickness if I can't relate it in any way to the phase change?
 
jegues said:
So the thickness doesn't relate to its phase change at all?
Sorry for the typo. The phase change is 2pi*x*n/lambda.

ehild
 
ehild said:
Sorry for the typo. The phase change is 2pi*x*n/lambda.

ehild

Wow, I actually managed to solve this one...

\phi_{1} = \frac{2\pi nx}{\lamda}

\phi_{2} = \frac{2\pi x}{\lamda}

\Delta \phi = \frac{2 \pi x (n-1)}{\lamda} = \pi

Therefore, 488nm.

Now I just got to figure my other problems out :wink:!
 

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