Thin Film Interference (Interferometer)

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SUMMARY

The forum discussion focuses on calculating the index of refraction of a gas using an interferometer setup with a cavity depth of 1.30 cm and a light wavelength of 610 nm. A total of 236 dark fringes were observed, leading to the equation 1.3(n-1) = 235.5(λ)/n. The calculated index of refraction values ranged from 1.001537 to 1.01093, suggesting potential candidates like Carbon disulphide or Ethyl ether. The discussion emphasizes the importance of understanding how the wavelength changes in different media and the role of phase shifts in interference patterns.

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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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