# Homework Help: Interferometer / Index of Refraction

1. Dec 9, 2009

### hm29168

1. The problem statement, all variables and given/known data

Giancoli Ed. 6, Ch. 24, #52

One of the beams of an interferometer passes through a small glass container containing a cavity 1.3 cm deep. When a gas is allowed to slowly fill the container, a total of 236 dark fringes are counted to move past a reference line. The light used has a wavelength of 610nm. Calculate the index of refraction of the gas, assuming that the interferometer is in a vacuum.

2. Relevant equations

For destructive interference:

Distance traveled = (m+0.5) * wavelength
where m is in the set 1, 2, 3....

3. The attempt at a solution

http://en.wikipedia.org/wiki/File:Interferometer.svg is a picture of an interferometer. The gas chamber is placed between the silvered mirror and the mirror on the right and the reflections are in line with each other instead of slightly at an angle in this case.

Let d be the difference in distance between the beam going through the gas and what the beam would be not going through the gas. Since the light passes twice through the gas chamber, the total difference in distance will be 2d. Plugging into our equation, we get

$$2d=(236+\frac{1}{2})\cdot(610\cdot10^{-9})$$
$$d=7.21\cdot10^{-5}$$

Armed with the distance, I should be able to find the index of refraction of the gas using Snell's law. However, I am not sure if I am calculating the extra distance traveled correctly. I also don't know how to find the index of refraction if the beam of light is in line with the normal of the medium.

Thanks for all of your help!

2. Dec 10, 2009

### Redbelly98

Staff Emeritus
Welcome to Physics Forums.
Snell's Law is not relevant here.

Note, the beam does not actually travel any extra distance. This is about figuring out the # of wavelengths that span the cavity length 2d (2d since the beam makes 2 passes through the cavity.

As a hint, think in terms of how the wavelength depends on n. When the cavity fills with gas, there are ____(how many?) more wavelengths spanning the distance 2d than when the cavity was under vacuum.

Another way to think about this: when the detector goes through 1 complete fringe cycle, i.e one dark fringe to the next dark fringe, that means one of the paths has either 1 more or 1 less wavelength in it. In this case it doesn't happen by changing the distance, but rather by changing the wavelength within the cavity being filled by the gas.

3. Dec 10, 2009

### hm29168

Thanks for the welcome.

Ok, so the wavelength of light in a medium that's not a vacuum can be found by
new wavelength = (wavelength in vacuum) / (index of refraction)

So, the gas will cause the beam to be out of phase compared to the other beam due to the temporary change in wavelength from its time in the gas. Since we know the distance that the beam travels (twice the length of the gas container) and the number of dark stripes, we can reuse our old equation and find the new wavelength so that we can find the index of refraction.

For destructive interference:
Distance traveled = (m+0.5) * wavelength
2*0.013 = (236 + 0.5) * new wavelength
new wavelength = 1.099 * 10^-4 m

Then from the equation mentioned above:
new wavelength = (wavelength in vacuum) / (index of refraction)
1.099 * 10^-4 = (610*10^-9) / (index)
index = 0.055

This isn't reasonable as this is less than the index of refraction for a vacuum. Did I misunderstand the concept?

4. Dec 10, 2009

### Redbelly98

Staff Emeritus
Okay, good.

Ah, not true. m is not the number of wavelengths in the 2d cell distance here; it is the number of additional wavelengths that are in the cell, after the gas is added, compared to when the cell was under vacuum.

5. Dec 18, 2009

### hm29168

I am really sorry for the late reply. Exams really put a stop to self-studying.

Ok, to find how many wavelengths there were in the cell originally under a vacuum, we can divide 2d by the wavelength of the light. So, in the new gas environment, the number of wavelengths is 2d divided by the new wavelength from refraction.

Initial number of wavelengths passing through: 2*0.013 / (610*10^-9) = 42623
Since there are 236 dark stripes passing, so adding this to the initial number of wavelengths passing through and solving for the wavelength, we get:
2*0.013 / (λ) = 42623+236
λ = 606 nm

new wavelength = (wavelength in vacuum) / (index of refraction)
606*10^-9 = (610*10^-9) / (index)
index = 0.99

This still isn't reasonable. I think I am not quite sure how to interpret the dark stripes. Since there is a light stripe in between the dark stripes, should we multiply the number by 2? Thanks again for your help.

Last edited: Dec 18, 2009
6. Dec 18, 2009

### Redbelly98

Staff Emeritus
Looks good so far.
An n of less than 1 should sound an alarm bell. Try solving the above equation again.
Also note, (610 nm)/0.99 is 616 nm, not 606.

No, each fringe spacing -- whether it's dark-to-dark or bright-to-bright -- corresponds to a full wavelength in the beam path.

7. Dec 19, 2009

### hm29168

Oops. Algebra mistake:
606*10^-9 = (610*10^-9) / (index)
606*(index) = 610
index = 610 / 606 = 1.01

Thank you so much for all of your help!