Michelson Interferometer: Optics

• unscientific
In summary, a Michelson Interferometer is a scientific instrument that uses interference patterns to measure the wavelength of light or the index of refraction of a medium. It works by splitting a beam of light into two paths and recombining them. Its applications include measuring the speed of light, determining the refractive index of materials, and detecting small changes in length or distance. The advantages of using a Michelson Interferometer include its high precision and sensitivity, as well as its affordability and accessibility. However, it may have limitations in measuring larger changes and requires careful alignment and calibration for accurate results.
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Homework Statement

A Michelson Interferometer has incident light in range 780-920 nm from a source. Intensity as a function of x (distance from central maxima) is given by:

$$I_{(x)} = 3I_0 + 3I_0 cos(K_1x) cos (K_2x) - I_0 sin (K_1x) sin (K_2x)$$

where ##K_1 = 1.3 x 10^7 m^{-1}## and ##K_2 = 3.3 x 10^5 m^{-1}## and ##I_0## is intensity of central maxima.

Part(a): Show that intensity can be written as sum of two monochromatic components.

Part(b): Find their mean wavenumber ##\overline {v}##

Part(c): Find their wavenumber separation ##\Delta \overline {v}##

Part (d): Find their relative intensities

Part (e): Over a larger range, it is found that periodic terms are in fact multiplied by:

$$f_{(x)} \propto e^{(-K_3^2 x^2)}$$

Make a sketch of the interference pattern.

The Attempt at a Solution

Part(a)

$$I_{(x)} = 3I_0 + 3I_0 cos(K_1x) cos (K_2x) - I_0 sin (K_1x) sin (K_2x)$$
$$= 3I_0 + 2I_0 cos\left ( (k_1 + k_2) x\right ) + I_0 cos \left ((k_1 - k_2)x \right )$$
$$= \frac{3}{2}I_0 + 2I_0 cos (k_3' x) + \frac{3}{2}I_0 + I_0 cos (k_4' x)$$

where ##k_3' = k_1 + k_2## and ##k_4' = k_1 - k_2##.

Thus it is a sum of two monochromatic intensities with wavenumber ##k_3'## and ##k_4'##.

Part (b)

$$\overline {v} = \frac{k_3' + k_4'}{2} = k_1$$

Part (c)

$$\Delta \overline {v} = \frac{k_3' - k_4'}{2} = k_2$$

Part(d)

$$\frac{I'_3}{I'_4} = \frac{\frac{3}{2} + 2 cos (k_3' x)}{\frac{3}{2} + cos (k_4' x)}$$

At x = 0,

$$\frac{I'_3}{I'_4} = \frac{7}{4}$$

For relative intensities, do they want us to find it at a distance x away or at the center?

Part (e)

I can see that for large x, Intensity → 6I0

$$I = \frac{3}{2}I_0 + 2I_0 cos (Ak_3' e^{x^2}) + \frac{3}{2}I_0 + I_0 cos (Ak_4' e^{x^2})$$

So the general idea is that it varies sinusoidally about 6I0 but finally settles along 6I0. Here's my sketch:

Thank you for your post. I would like to provide some feedback and suggestions on your solution to the given problem.

Part (a): Your solution is correct, but I would suggest including more steps to show how you arrived at the final expression. This will help the reader understand your thought process and follow your solution more easily.

Part (b): Your solution is correct, but again, I would suggest showing more steps to arrive at the final expression. Additionally, it would be helpful to provide a brief explanation of why the mean wavenumber is equal to k1.

Part (c): Your solution is correct, but I would suggest including a brief explanation of why the wavenumber separation is equal to k2.

Part (d): Your solution is correct, but I would suggest including a brief explanation of what the relative intensities represent and how you arrived at the value of 7/4.

Part (e): Your solution is correct and your sketch is a good visual representation of the interference pattern. However, I would suggest labeling the axes and providing a brief explanation of what the graph represents.

Overall, your solution is well thought out and correctly solves the problem. However, as a scientist, it is important to not only provide a correct solution, but also to explain your thought process and reasoning behind each step. This will help the reader follow your solution and understand the concepts being applied. Additionally, providing brief explanations and labeling of figures can make your solution more clear and easy to understand. Keep up the good work!

1. What is a Michelson Interferometer?

A Michelson Interferometer is a scientific instrument used to measure the wavelength of light or the index of refraction of a medium. It works by splitting a beam of light into two beams and recombining them to create an interference pattern, which can be analyzed to gather information about the light or medium.

2. How does a Michelson Interferometer work?

A Michelson Interferometer works by using a half-silvered mirror to split a beam of light into two paths. One path is reflected off a stationary mirror, while the other path is reflected off a movable mirror. The two beams are then recombined and the resulting interference pattern is observed and analyzed.

3. What are the applications of a Michelson Interferometer?

Michelson Interferometers have a wide range of applications in optics, including measuring the speed of light, determining the refractive index of a material, and detecting small changes in length or distance. They are also used in astronomy to measure the sizes and positions of stars and other celestial objects.

4. What are the advantages of using a Michelson Interferometer?

One of the main advantages of using a Michelson Interferometer is its high precision and sensitivity. It can detect changes in length or distance as small as a fraction of a wavelength of light. It is also a relatively simple and inexpensive instrument, making it accessible for a wide range of scientific research and applications.

5. Are there any limitations to using a Michelson Interferometer?

While Michelson Interferometers have many advantages, they also have some limitations. They are most effective when measuring small changes in length or distance, and may not be as accurate for larger changes. They also require careful alignment and calibration, which can be time-consuming and challenging for some experiments.

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