White light interference pattern and its resolution criteria

[Clara Chung]

Homework Statement

Homework Equations

The Attempt at a Solution

I don't know how to do part d.
For di, i know that separation of arm d should be small in order to have spatial coherence.
For dii, what is the meaningful criterion, can I say the primary max of blue light lies on the first zero of red light?
I have no idea of part iii too. Please help.

 

[Charles Link]

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.

 

[Clara Chung]

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.

Thank you so much for the large amount of information. I will process it and try.

 

[Charles Link]

@Clara Chung What might be the most important part in analyzing the white light interference is the equation, $I(\theta)=I_o \cos^2( \frac{\pi D \, cos(\theta)}{\lambda})$ from post 7 of this "link" : https://www.physicsforums.com/threa...michelson-interferometer.933638/#post-5902650 That equation is derived in post 8 of the same "link". $\\$ Additional note: It's been a while since I looked at this posting in the "link", so I had to study it a little: The distance $D$ in this posting is defined as the path length difference in distance (on-axis) $D=2d$. $\\$ Notice you will get interference maxima (the brightest part of the interference fringe/ring) when $D \cos(\theta)=2d \cos(\theta)=m \lambda$.