Interference and Diffraction

[Aphrael]

Homework Statement

The distance between the movable mirror and the beam splitter in a Michelson interferometer is increased a small amount. What this happens, you see 200 dark fringes move across the field of view. If the incident light was 600nm, by how much was the mirror moved (in millimeters)?

Homework Equations

I am not sure at all how to begin this problem. I do not think that I can use the equation sin (theta) =m(wavelength)/d because the only variables I know is wavelength. I am not sure what to do with the 200 dark fringes because I do not think that m=200. I also do not know if I could use x =nLw/d because although I think I am solving for L, I do not know x,n, or d. How do I begin?

 

[anathema]

Thank you for your question. To solve this problem, we can use the equation for the path difference in a Michelson interferometer:

Δx = 2dΔm

Where Δx is the change in path difference, d is the distance between the movable mirror and the beam splitter, and Δm is the number of dark fringes moved.

We know that the wavelength of the incident light is 600nm, so we can plug that into the equation:

Δx = 2dΔm = 2(600nm)Δm

Now, we are given that 200 dark fringes have moved across the field of view. This means that Δm = 200. Plugging this into the equation, we get:

Δx = 2(600nm)(200) = 240,000nm

However, we are asked to find the distance in millimeters, so we need to convert the units. 1nm = 10^-9m, and 1m = 1000mm, so 1nm = 10^-6mm. Therefore,:

Δx = 240,000nm = 240,000(10^-6)mm = 0.24mm

So, the mirror was moved 0.24mm.

I hope this helps. Let me know if you have any further questions.

 

[Moetasim]

To solve this problem, we can use the equation for the number of fringes, N, in an interference pattern: N = d/lambda, where d is the distance between the movable mirror and the beam splitter, and lambda is the wavelength of the incident light. Since we are given the number of fringes (200) and the wavelength (600nm), we can rearrange the equation to solve for d: d = N*lambda.

Therefore, the mirror was moved a distance of 200*600nm = 120,000nm or 0.12mm.

It is important to note that the equation sin(theta) = m*lambda/d is used for diffraction patterns, not interference patterns. In this case, we are dealing with an interference pattern, so the equation for the number of fringes is more appropriate.

I hope this helps clarify the problem for you. If you have any further questions, please don't hesitate to ask.