Thin Film Interference (Interferometer)

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Homework Help Overview

The problem involves an interferometer where a beam passes through a gas-filled cavity, and participants are tasked with determining the index of refraction of the gas based on the observed dark fringes and the wavelength of light used.

Discussion Character

  • Exploratory, Assumption checking, Conceptual clarification

Approaches and Questions Raised

  • Participants discuss the relationship between the number of dark fringes observed and the extra distance traveled by light in the gas. There are questions about the meaning of 'new' wavelength and the role of the glass container. Some participants attempt to derive the index of refraction using different approaches, including time calculations and wavelength considerations.

Discussion Status

The discussion is ongoing, with various interpretations being explored. Some participants have provided insights into the calculations and assumptions involved, while others express uncertainty about the values obtained and the implications of the results.

Contextual Notes

Participants note potential discrepancies in the calculated index of refraction, with some values suggesting unrealistic gas properties. The discussion also reflects on the assumptions made regarding the wavelength and phase changes during the interference process.

turdferguson
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Homework Statement


One of the beams of an interferometer passes through a small glass container containing a cavity 1.30 cm deep. When a gas is allowed to slowly fill thr container, a total of 236 dark fringes are counted to move past a reference line. The light used has a wavelength of 610 nm. Calculate the index of refraction of the gas, assuming the interferometer is in a vacuum


Homework Equations


extra distance = m*lambda/n = twice the depth

d=vt ??

The Attempt at a Solution


The first dark spot occurs when the extra distance is half the new lambda. This means the 236th dark spot occurs when m = 235.5 I attempted to solve for n by equating this to twice the depth, but got something way lower than 1. I must be missing something big. Does the glass container outside play a role?
 
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I'm not sure what you mean by 'new' lambda. The wavelength does not change, only the phase, surely ? If the lines are displaced by 236*lambda/2 then the final phase change must equal this. Though, I must admit it seems a large value for 1.3 cm of gas.
 
As a beam of light moves into a material of higher index of refraction, its speed slows down to c/n and its wavelength shortens to lambda/n. Frequency is held constant
 
Thank you, I'm sorry if I wasted your time.

I think the extra distance you're after is 1.3(n-1) which I got by working out how much longer it takes to get through and multiplying by c.
 
Mentz114 said:
Thank you, I'm sorry if I wasted your time.

I think the extra distance you're after is 1.3(n-1) which I got by working out how much longer it takes to get through and multiplying by c.
I don't follow, can you explain how you determined the time?
 
I get n = 1.001537. Could be Carbon disulphide or Ethyl ether

But I could also be wrong.

[I've just seen your post]

t1 = 1.30/c
t2= 1.30/(c/n)

t2-t1 = 1.3n/c - 1.3/c

(t2-t1)*c = 1.3(n-1)
 
Ok, that makes sense. But I used 1.3(n-1)=235.5(lambda)/n and got a quadratic with a root at 1.01093 Thanks for the help
 
Well, I think the interference is at the unslowed wavelength, so your 1/n factor on the right isn't needed. Also, your value is much higher than any real gas ( that I know of).

Glad to be be of some help.
 
Last edited:

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