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  5. The michelson interferometer

3.5 The michelson interferometer  (Page 2/7)

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Measuring the refractive index of a gas

In one arm of a Michelson interferometer, a glass chamber is placed with attachments for evacuating the inside and putting gases in it. The space inside the container is 2 cm wide. Initially, the container is empty. As gas is slowly let into the chamber, you observe that dark fringes move past a reference line in the field of observation. By the time the chamber is filled to the desired pressure, you have counted 122 fringes move past the reference line. The wavelength of the light used is 632.8 nm. What is the refractive index of this gas?

Pictures shows a schematic of a set-up utilized to measure the refractive index of a gas. The glass chamber with a gas is placed in the Michelson interferometer between the half-silvered mirror M and mirror M1. The space inside the container is 2 cm wide.

Strategy

The m = 122 fringes observed compose the difference between the number of wavelengths that fit within the empty chamber (vacuum) and the number of wavelengths that fit within the same chamber when it is gas-filled. The wavelength in the filled chamber is shorter by a factor of n , the index of refraction.

Solution

The ray travels a distance t = 2 cm to the right through the glass chamber and another distance t to the left upon reflection. The total travel is L = 2 t . When empty, the number of wavelengths that fit in this chamber is

N 0 = L λ 0 = 2 t λ 0

where λ 0 = 632.8 nm is the wavelength in vacuum of the light used. In any other medium, the wavelength is λ = λ 0 / n and the number of wavelengths that fit in the gas-filled chamber is

N = L λ = 2 t λ 0 / n .

The number of fringes observed in the transition is

m = N − N 0 , = 2 t λ 0 / n − 2 t λ 0 , = 2 t λ 0 ( n − 1 ) .

Solving for ( n − 1 ) gives

n − 1 = m ( λ 0 2 t ) = 122 ( 632.8 × 10 −9 m 2 ( 2 × 10 −2 m ) ) = 0.0019

and n = 1.0019 .

Significance

The indices of refraction for gases are so close to that of vacuum, that we normally consider them equal to 1. The difference between 1 and 1.0019 is so small that measuring it requires a correspondingly sensitive technique such as interferometry. We cannot, for example, hope to measure this value using techniques based simply on Snell’s law.

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Check Your Understanding Although m , the number of fringes observed, is an integer, which is often regarded as having zero uncertainty, in practical terms, it is all too easy to lose track when counting fringes. In [link] , if you estimate that you might have missed as many as five fringes when you reported m = 122 fringes, (a) is the value for the index of refraction worked out in [link] too large or too small? (b) By how much?

a. too small; b. up to 8 × 10 −5

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Problem-solving strategy: wave optics

Step 1. Examine the situation to determine that interference is involved . Identify whether slits, thin films, or interferometers are considered in the problem.

Step 2. If slits are involved , note that diffraction gratings and double slits produce very similar interference patterns, but that gratings have narrower (sharper) maxima. Single-slit patterns are characterized by a large central maximum and smaller maxima to the sides.

Step 3. If thin-film interference or an interferometer is involved, take note of the path length difference between the two rays that interfere . Be certain to use the wavelength in the medium involved, since it differs from the wavelength in vacuum. Note also that there is an additional λ / 2 phase shift when light reflects from a medium with a greater index of refraction.

Step 4. Identify exactly what needs to be determined in the problem (identify the unknowns) . A written list is useful. Draw a diagram of the situation. Labeling the diagram is useful.

Step 5. Make a list of what is given or can be inferred from the problem as stated (identify the knowns) .

Step 6. Solve the appropriate equation for the quantity to be determined (the unknown) and enter the knowns . Slits, gratings, and the Rayleigh limit involve equations.

Step 7. For thin-film interference, you have constructive interference for a total shift that is an integral number of wavelengths. You have destructive interference for a total shift of a half-integral number of wavelengths . Always keep in mind that crest to crest is constructive whereas crest to trough is destructive.

Step 8. Check to see if the answer is reasonable: Does it make sense? Angles in interference patterns cannot be greater than 90 ° , for example.
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Source:  OpenStax, University physics volume 3. OpenStax CNX. Nov 04, 2016 Download for free at http://cnx.org/content/col12067/1.4
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