# Thin Film Interference (Interferometer)

• turdferguson
In summary, when a gas is allowed to slowly fill a small glass container, a total of 236 dark fringes are counted. The index of refraction of the gas is calculated to be 1.30.f

## 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?

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.

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.

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.

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