Part (d) especially is asking you to answer a rather specialized question without giving much detail on the particular arrangement. There are two types of Michelson configurations=plane wave and diffuse source. The plane wave form results in the entire screen seeing an intensity that depends on the interference. For the diffuse source type, you get a series of rings. See the following "link" post 3:
https://www.physicsforums.com/threa...michelson-interferometer.933638/#post-5902650 ## \\ ## In my opinion, [Edit: The "white light" interference, upon working out the details, (and seeing that you won't get a true rainbow effect), appears to be of somewhat secondary significance =the detail of the "white light' interference was previously omitted in my classwork and laboratory schooling that I had using the Michelson interferometer], they would do well to present more material, and quiz you on that. It appears you are kind of on the right track though=for white light, the spacing ## d ## needs to be very small, really basically ## d=0 ## to get white light to interfere. ## d=0 ## should get you an ## m=0 ## bright spot, independent of ## \theta ##, (where ## m \lambda=2d \cos(\theta) ##), that will occur regardless of wavelength. ( Let's assume this is a diffuse source type Michelson). The rings will be most prominent for white light at ## m=1 ##, just a little away from ## d=0 ##, if there happens to be more of one selected color in the spectrum=e.g. more blue than red. ## \\ ## It is rather difficult to quantify what really appears to be a rather vague question that they are giving you. You could try starting with just blue light for ## d=1 ##, and mapping out a ring pattern, and overlaying a red light ring pattern to see what you get. See also post 7 of the "link" above for the equation that will quantify the intensity of the ring pattern as a function of ## d ##, ## \lambda ##, and ## \theta ##. ## \\ ## Editing: I have one additional idea: Right around ## m=1, ## I think the observed pattern may look like the colors of the rainbow [Edit: See below=this part here is somewhat in error], with the shorter wavelengths constructively interfering for larger ## \theta ## than the longer wavelengths. There will start to be appreciably mixing when ## 2 \lambda_{blue}=2d ## i.e. when ## m=2 ## for blue, because then you have very nearly ## m=1 ## for red at the same angle as ## m=2 ## for blue. (Not precisely, but I'll let you fill in the details). I'll let you determine if there still may be some interference present for this case. ## \\ ## Additional editing: With a monochromatic source the interference effect is fringes that vary in intensity. With a white light source, the interference effect is a rainbow pattern= [Edit: See below: This is somewhat in error] with the intensity fairly uniform over the entire pattern. ## \\ ## And one more input for this problem: If the output of the Michelson is focused onto a screen for viewing, ## \theta ## may be somewhat limited, and you might have a span of ## \theta ## something like ## 0<\theta <30^o ##. Thereby, at ## m=1 ##, you probably won't see all of the colors of the rainbow at the same time at one single position ##d ##. ## \\ ## And one part needs correcting here: It will be the colors of the rainbow with much overlap: Unlike a prism or diffraction grating where ## \theta=\theta(\lambda) ##, here each ## \lambda ## has a wide fringe pattern and is not precisely located like it is with a prism or diffraction grating. My previous statement, that you see the colors off the rainbow is somewhat in error, but I am going to leave it there, so you can see where it is incorrect.
