# Interferometer / Index of Refraction

• hm29168
In summary: Distance traveled = (m+0.5) * wavelength2d = (m+0.5) * wavelength2d = (m+0.5) * (wavelength in vacuum) / (index of refraction)2d = (m+0.5) * wavelength in vacuum * index of refraction2d = (m+0.5) * (610*10^-9) * index of refractionThen, we can do the same for when the gas chamber is empty, except we know that the number of dark fringes is 0.0
hm29168

## Homework Statement

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.

## Homework Equations

For destructive interference:

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

Welcome to Physics Forums.
hm29168 said:

## Homework Equations

For destructive interference:

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

Redbelly98 said:
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.

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?

hm29168 said:
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)
Okay, good.

For destructive interference:
Distance traveled = (m+0.5) * wavelength
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.

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

Redbelly98 said:
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.

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:
hm29168 said:
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)
Looks good so far.
index = 0.99
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.

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

Redbelly98 said:
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.

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!

## 1. What is an interferometer?

An interferometer is a scientific instrument used to measure the interference patterns created by the combination of two or more light waves. It is commonly used in physics and astronomy to measure small changes in distance or wavelength.

## 2. How does an interferometer work?

An interferometer works by splitting a beam of light into two or more paths, then recombining the beams to create an interference pattern. This pattern is then analyzed to measure changes in wavelength or distance.

## 3. What is the index of refraction?

The index of refraction is a measurement of how much a material can bend light as it passes through it. It is defined as the ratio of the speed of light in a vacuum to the speed of light in the material.

## 4. How is the index of refraction measured?

The index of refraction can be measured using an interferometer. By measuring the changes in the interference pattern as light passes through a material, scientists can calculate the index of refraction for that material.

## 5. What is the importance of the index of refraction in science?

The index of refraction is important in many fields of science, such as optics, astronomy, and material science. It helps us understand how light interacts with different materials and is crucial in the design of lenses, prisms, and other optical instruments.

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