# Homework Help: Fringe Visibility of Sodium Doublet in Michelson interferometer

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1. Dec 6, 2017

### Marcus95

1. The problem statement, all variables and given/known data
The sodium D-lines are a pair of narrow, closely spaced, approximately equal intensity spectral lines with a mean wavelength of approximately 589 nm. A Michelson interferometer is set up to study the D-lines from a sodium lamp. High contrast fringes are seen for zero pathlength difference between the two arms of the interferometer. The fringes disappear when the pathlength difference is increased to 0.29 mm.
a. What is the wavelength difference between the lines?
b. What would you expect to see if the pathlength difference were increased to 0.58 mm, assuming the spectral lines are very narrow?
c. If the spectral lines have approximately Gaussian shapes, with a width of 50 pm (taken between the points of the line shape where the intensity falls to e −1/2 of the peak intensity), what is the maximum fringe contrast (visibility) seen for a pathlength difference of around 4 mm?

2. Relevant equations
Fringe visibility: $v = \frac{I_{max}-I_{min}}{I_{max}+I_{min}}$

3. The attempt at a solution
a) The phase difference of the two beams increase by $\pi$ due to the pathlength difference, hence:
$\pi = 2\pi \delta x (\frac{1}{\lambda - \delta\lambda /2} - \frac{1}{\lambda + \delta\lambda /2} )$
giving $\delta\lambda = 0.60nm$

b) We now have $\delta x' = 2 \delta x$, hence $\delta \phi = 2\pi$ so the fringe pattern should be the same as initally.

c) This is the part where I get stuck. $\delta x'' = 4 mm$ gives $\delta \phi '' \approx 13.8\pi$, meaning that the peaks neither exactly overlap or exactly overlap with each others minima. However, from this point on I am not sure how to find their separation in the "line shape". What is the "line shape" plotted as a function of? Considering that the width of the peaks in the line shape is given in units pm, I suppose it must be some sort of length. However, I am not sure what length, because the one on the screen is not general enough (it depends on the distance etc.).

Could somebody please give me some guidance on how to attack the last part of the problem? Thank you. :)

2. Dec 6, 2017

You are missing the basic equation that you need to find the (polar) angle $\theta$ in the far field pattern (which you can also observe directly by focusing your eye at $+ \infty$ and looking into the interferometer) for which bright rings (in the case of $\theta=0$ it's a bright spot) occur in a Michelson interferometer: $m \lambda=2d \cos{\theta}$. ($m=$ integer ). The center bright spot at $m=0$ for $d=0$ is independent of wavelength. The center of dark rings occurs at angles $\theta$ given by $(m +1/2) \lambda=2d \cos{\theta}$. For the reading at $d=.29$ mm, the bright spot or ring of one wavelength matches up with the dark spot or ring from the other wavelength so that little or no fringes are seen. $\\$ I can give you an additional hint on this one: $m$, call it $m_o$, is going to be quite large for this case at $d=.29 \, mm$), and is given by $m_o \lambda_1=2 (.29 \, mm)$ where you use $\theta=0$, ($\cos{\theta}=1$). For the other wavelength, $\lambda _2=\lambda_1 - \Delta \lambda$, (I'm letting $\lambda_2$ be the shorter one), you get destructive interference for the same $m_o$ at $\theta=0$ for $d=.29 \, mm$. They essentially give you $\lambda_1$, and you need to find $\Delta \lambda$.

Last edited: Dec 6, 2017
3. Dec 6, 2017

Additional comment: The form of the Michelson interferometer that is used for the above experiment uses a diffuse source instead of a plane wave source. Any point on this diffuse source results in two point source images spaced by distance $2d$ apart. These point source images create the ring interference pattern in the far field, regardless of their location on the diffuser plate. The same angular ring interference pattern occurs in the far field for every point source that originates from the diffuser plate, with each point source having two separate but mutually coherent point source images because of the beamsplitter and two mirrors. It should be somewhat easy to show that $m \lambda =2d \cos{\theta}$ is the condition for constructive interference between the two point source images in the far field.

Last edited: Dec 6, 2017
4. Dec 11, 2017

### Marcus95

Thank you for your reply @Charles Link , although I am not sure I really follow. The equations you use look like the ones for thin film interference. Is $\theta$ the angle between a light ray and the perpendicular distance to the screen after the beam splitter? But if rays of the two wavelengths take the same path, why does $\cos\theta$ matter and where do we get the factor of 2 from?
this is not quite clear to me, how can two different images interfere?

Regarding the last part of the question, I still don't have any clue what the line-shape is or how this is to be used to find the fringe separation. Do you have any hint about that? Thank you!

5. Dec 11, 2017

This one is similar to thin film interference, but in this "diffuse source type" Michelson interferometer, the point sources replace the plane waves. The beamsplitter in a Michelson interferometer is used twice=first time is to split the original beam. The original beam for this diffuse source type Michelson interferometer can be thought of as a single point source from the diffuser plate. It radiates spherically. The beamsplitter sends part of the beam up to one mirror, and the other part of the beam to the other mirror. The part of the beam that goes straight through the beamsplitter is one point source, and its mirror image in the beamsplitter for the "partially reflected part of the original source by the beamsplitter" becomes a mirror image type point source for this problem. These two point sources take different paths (to the mirror and back to the beamsplitter, so the path lengths can be different), and the beamsplitter (partially) recombines them on their second encounter with the beamsplitter. Whether the intensity that is observed at some angle from these two point sources depends on whether they constructively interfere. $\\$ (I wish I had the luxury of a chalkboard for the explanation, because it is really somewhat simpler than these verbal explanations make it). $\\$ Important calculation: For a simple case: Take two point sources a distance $2d$ apart on the z-axis: You can determine their interference in the far-field in a plane parallel to the x-y plane. The interference will be a ring pattern. Result will be constructive interference for polar angle $\theta$, making a ring pattern with bright rings at $m \lambda=2d \cos{\theta }$.The same thing happens in this "diffuse source" version of the Michelson interferometer. The ring pattern will be the same in the far field, whether the source originates at the origin or a half-inch or an inch from the center of the diffuse plate. The ring patterns from all the pairs of point sources superimpose to make one bright pattern of rings.$\\$ There is also a plane wave version of the Michelson interferometer. This one works very much like thin film interference. I wrote about it in detail in an Insights article about a year ago. https://www.physicsforums.com/insights/fabry-perot-michelson-interferometry-fundamental-approach/ You might find it of interest, but one feedback that I received is that it needs more diagrams. Again, if we had the luxury of a chalkboard, the explanations would probably be easier to follow. This plane wave version is also really somewhat simple, but diagrams could add much to the verbal description. $\\$ Additional comment: The Michelson interferometer, in both versions, diffuse source and plane wave, is really a specialized topic, so it is usually taught in some detail in some of the Optics textbooks, but I don't know of any that does a really good job with either version. The plane wave version is one that took me a long time to figure out=the antireflection coating on one face of the beamsplitter is an important feature. In addition, I hadn't seen it explained using the Fresnel coefficients in any textbook. That is something I was finally able to piece together. The interference that occurs between two beams incident on a single beamsplitter from opposite directions is quite remarkable. Again, for additional details on the plane wave version, see my Insights article with the above "link".

Last edited: Dec 11, 2017
6. Dec 14, 2017

### Marcus95

Thank you @Charles Link , I believe I have gathered a much better insight to Michelson-Morley Inferometers now! :)
However, I am afraid I still don't know how to approach the last part of the question I am facing. What is meant by the line-shape and how do I relate the width of it to the fringe contrast?

7. Dec 14, 2017

That one (the fringe contrast) is not an easy question, even though I think the theoretical understanding of the Michelson interferometer is complete enough to be able to quantify the results by using the Fresnel coefficient method. For a single monochomatic wavelength, I believe the intensity pattern is basically $\\$ (editing): $I(\theta)=I_o cos^2(\frac{ \pi D \cos{\theta}}{\lambda}+\phi_o )$ for some $\phi_o$ . (I'm working somewhat on my instincts here=I haven't completely proven this result, but I'm pretty sure that's what you get if the fresnel coefficients $\rho$ and $\tau$ are such that $\rho=\pm 1/\sqrt{2}$ and $\tau=1/\sqrt{2}$ from an ideal 50-50 beamsplitter). $\\$ It would make it easier to work with just a single wavelength from the sodium doublet and assign a spectral linewidth of $\Delta \lambda=.050$ nm. The intensity number $I_{max}$ and $I_{min}$ could be found by numerical integration of the Gaussian lineshape, assigning a $\phi_o$ to each element of the spectrum $I(\lambda) \, d \lambda$ and determining the matching intensity pattern $I(\theta)$. You basically need the intensity ($I_{max}$) at $\theta=0$, and for $\theta$ such that you have ( $I_{min}$) the center of the first dark ring next to the bright spot at $\theta=0$, with the distance chosen to be $4 \, mm$ plus any additional $\Delta$ that may be necessary to center/maximize the bright spot at $\lambda=\lambda_o$ where $\lambda_o$ is one of the wavelengths of the doublet. (You can also assume that the other wavelength of the doublet has a pattern that matches this one=something that you solved for in (a) and (b)). Perhaps even a closed form integration could be done for these under certain Taylor series approximations. This again, is not an easy question, but would be quite workable if you want to spend a couple of hours on it. $\\$ Basically (editing) $\phi_o= -\frac{ \pi D}{\lambda_0 }$, with $D=4 \, mm$,(path distance difference=2x 2 mm), and with a Gaussian spectral line: $I(\lambda)=I_o e^{-(\lambda-\lambda_o)^2/(\Delta_w)^2}$, and $\lambda$ needs to get integrated basically from $-\infty$ to $+\infty$ (an approximation because wavelength $\lambda$ is never negative). (And I don't have the complete linewidth equation yet, because $\Delta_w$ needs to be determined. They want to define the linewidth by $e^{-1/2}$ of the peak intensity=perhaps somewhat fussy=they could do well to provide you width the line intensity equation). They seem to be saying at $|\lambda-\lambda_o|=.050/2 \, nm$, that $I/I_o=e^{-1/2}$. With a little algebra, you could determine what $\Delta_w$ needs to be in my spectral equation. (Editing: I get $\Delta_w=.050/ \sqrt{2} \, nm$ if my algebra is correct.) (Additional note: This $I_o$ is different from the previous $I_o$, but for this problem, all factors of $I_o$ could be set equal to $1$, since we are simply determining a contrast ratio). $\\$ Just a couple additional calculations that might be helpful : $D=(m+1/2) \lambda_o$ (for constructive interference at $\theta=0$), and $D \cos(\theta)=m \lambda_o$ (for a dark ring for some angle $\theta$). I get $\frac{ \pi D}{\lambda_o} (\cos(\theta)-1)=-\pi/2$. This means $1-\cos(\theta)=\lambda_o /(2 D )$ if my algebra is correct, and with $\cos(\theta) \approx 1-\theta^2/2$, this means $\theta \approx \sqrt{\lambda_o /D}$ is $\theta$ where you need to compute the intensity $I_{min}$, along with $I_{max}$ at $\theta=0$. (The $1/2$ in the equation for constructive interference at $\theta=0$ comes from a $\pi$ phase change that occurs for the external reflection from the beamsplitter. Similarly, this 1/2 is absent from the equation for destructive interference). $\\$ Additional item: To be more exact, the intensity equation is $I(\theta)=I_o \cos^2(\frac{ \pi D \cos{\theta}}{\lambda}+\frac{\pi}{2})$. In the slightly modified form above $D$ is rewritten as $D=D-\Delta$, where $\Delta$ is a very small distance that will cause the peak at $\theta =0$ to occur at $D$. In addition, in choosing $\phi_o$ how we did, we had the luxury of subtracting $m \pi$ from the complete term contained in the phase of the $cos^2(\theta+\phi_o )$ term. Thereby, the intensity equation takes on the form provided above. $\\$ In summary, this part of the problem is rather non-elementary, but if you were to now do a numerical computation of it, I think it would generate a sensible result for $I_{max}$ and $I_{min}$, from which you could compute the fringe contrast.

Last edited: Dec 14, 2017
8. Dec 14, 2017

And I spotted one important error in my above post: Starting with $E=E_o [\cos(\omega t)+\cos(\omega t +\frac{2 \pi D \cos{\theta}}{\lambda})]=2E_o \cos(\omega t +\frac{\pi D \cos{\theta}}{\lambda}) \cos(\frac{\pi D \cos{\theta}}{\lambda})$ (by a trigonometric identity), the result is $I(\theta) =proportional \, to \,E^2 \, without \, the \, time \, dependence \,=I_o \cos^2 (\frac{\pi D \cos{\theta}}{\lambda} )$. $\\$ (Originally I incorrectly had $I(\theta)=I_o \cos^2(\frac{2 \pi D \cos{\theta}}{\lambda}+\phi_o )$, etc. in the above equations. I have now made the necessary corrections). $\\$ Additional note: In this last simple computation of the intensity, I have omitted the $\pi$ phase change that occurs on one of the reflections from the beamsplitter. With this $\pi$ term included, the $\frac{\pi}{2}$ phase term appears in the final exact equation for $I(\theta)$. $\\$ Additional item: With an approximation or two, along with a couple of algebraic substitutions, $I_{max}$ and $I_{min}$ were readily evaluated in closed form. I was able to show, (I haven't double-checked every detail of the result, but I think it is correct), that $v=\frac{I_{max}-I_{min}}{I_{max}+I_{min}}=e^{-b^2}$ where $b=\frac{\pi D \Delta_w}{\lambda_o^2}$. $\\$ If you @Marcus95 were able to follow the results up to this point, I'd be happy to lead you through the details of this calculation. Here are some of the details: $I_{max}=(1/2)+(1/2)e^{-b^2}$ and $I_{min}=(1/2)-(1/2)e^{-b^2}$, which thereby resulted in the expression for visibility $v$. To derive these results, I used a properly normalized Gaussian spectral intensity function $I(\lambda)=\frac{1}{\sqrt{\pi} \Delta_w} e^{-(\lambda-\lambda_o)^2/\Delta_w^2}$. One integral that resulted was $\int\limits_{-\infty}^{+\infty} cos^2(bx)e^{-x^2} \, dx$. Writing $cos^2(bx)=(cos(2bx)+1)/2$ allowed for the complete evaluation. (I was able to google the result for $\int\limits_{0}^{+\infty} cos(ax)e^{-x^2} \, dx$ ).$\\$ For the case of $1-cos(\theta)=\lambda_o/(2 D)$, the result was $cos^2(bx+\pi/2)$ in the $cos^2$ term in the integral. One approximation that was used is $\frac{\pi D}{\lambda}=\frac{\pi D}{(\lambda_o+\Delta \lambda)} \approx \frac{\pi D}{\lambda_o}(1-\frac{\Delta \lambda}{\lambda_o})$. $\\$ This last result of the fringe visibility is a quantitative result of how the fringes wash out in a Michelson interferometer for large $D$ if the source doesn't have the necessary coherence length, corresponding to a sufficiently narrow line width $\Delta_w$. It is possible this result could be googled somewhere, but we succeeded in deriving it. And it took a little effort to crank out the result, but the calculation is really rather straightforward, and just requires some knowledge of calculus.

Last edited: Dec 15, 2017
9. Jan 3, 2018

@Marcus95 This experiment involving the interference fringes of the sodium doublet with a Michelson interferometer is performed on occasion in Optics courses that include laboratory experiments. My classmates and I performed this same experiment in the upper level undergraduate Optics course at the University of Illinois at Urbana in 1976. $\\$ Our experimental results did show, in fact, that the lines of the doublet are separated by $\Delta \lambda \approx 6.0$ Angstroms, by finding the distance between two successive $D's$ where the interference fringes matched up for the two wavelengths as described in part (a).