Strategy Refer to Figure for the geometry. We use the result of the Michelson interferometer interference condition to find the distance moved,. Solution For a nm red laser light, and for each fringe crossing , the distance traveled by if you keep fixed is.
Significance An important application of this measurement is the definition of the standard meter. As mentioned in Units and Measurement , the length of the standard meter was once defined as the mirror displacement in a Michelson interferometer corresponding to 1,, 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 fringes move past the reference line. The wavelength of the light used is What is the refractive index of this gas? Strategy The 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 to the right through the glass chamber and another distance t to the left upon reflection. The total travel is. When empty, the number of wavelengths that fit in this chamber is. In any other medium, the wavelength is and the number of wavelengths that fit in the gas-filled chamber is.
Solving for gives. 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. 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 Figure , if you estimate that you might have missed as many as five fringes when you reported fringes, a is the value for the index of refraction worked out in Figure too large or too small?
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 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 , for example. Describe how a Michelson interferometer can be used to measure the index of refraction of a gas including air. In one arm, place a transparent chamber to be filled with the gas. See Figure. A Michelson interferometer has two equal arms. A mercury light of wavelength nm is used for the interferometer and stable fringes are found. One of the arms is moved by. How many fringes will cross the observing field? What is the distance moved by the traveling mirror of a Michelson interferometer that corresponds to fringes passing by a point of the observation screen?
Assume that the interferometer is illuminated with a nm spectral line of krypton When the traveling mirror of a Michelson interferometer is moved , 90 fringes pass by a point on the observation screen.
What is the wavelength of the light used? In a Michelson interferometer, light of wavelength When one of the mirrors is moved by a distance D , 8 fringes move past the field of view.
What is the value of the distance D? A chamber 5. The light used has a wavelength of nm in a vacuum. While all the air is being pumped out of the chamber, 29 fringes pass by a point on the observation screen. What is the refractive index of the air? Additional Problems For nm wavelength light and a slit separation of 0.
If the light source in the preceding problem is changed, the angular position of the third maximum is found to be. Interference in a Michelson interferometer can be understood in terms of thin-film interference. Imagine that the arms of the interferometer are rotated, such that there is a single optical axis, as shown in the drawing below. Then reflection from the mirrors M 1 and M 2 is analogous to reflection from two surfaces of an air gap of thickness d.
Since the phase shift upon reflection is the same for both mirrors, we find. If the two mirrors are precisely aligned such that their planes are exactly perpendicular to one another, thus ensuring that path differences over different regions of the mirrors are constant, the fringe pattern will consist of a series of concentric rings.
Each ring will correspond to a different angle of view, measured from the normal to the mirror M 1. These fringes are called fringes of equal inclination.
Your first statement is not completely accurate. M1 and M2 do need to be equal distance or equal full wavelengths from the beam splitter in order to get interference and create the rings but the final distance to the screen will also determine if the center is solid dark or light. Q1-The two recombined rays are projected onto a screen.
You do not want to look directly into the on coming rays. Q2-The thickness of the rings get smaller as you move away from the center. The number of rings doesn't change but the number that you see depends on the distance, how its focused and the overall lighting or intensity.
You can focus on a few in the center or hundreds with each one getting thinner and thinner as you move out. Sign up to join this community. The best answers are voted up and rise to the top. Stack Overflow for Teams — Collaborate and share knowledge with a private group. Create a free Team What is Teams?
Learn more. Michelson interferometer circular fringes Ask Question. Asked 4 years, 8 months ago. Active 1 year, 5 months ago. Viewed 28k times. Improve this question. Add a comment. Active Oldest Votes. Update as a result of a comment Schematic diagrams greatly exaggerated. Improve this answer. Farcher Farcher Does this make sense? You cannot get it from my diagram as it does not have the beam splitter in it.
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