Interferometer- 2 spectral lines

[Physgeek64]

Homework Statement

An interferometer is illuminated by light from a sodium lamp, which emits
two narrow spectral lines at wavelengths of 589.0nm and 589.6nm, with the intensity
of the 589.0nm line being twice that of the 589.6nm line. Show that there are values
of d at which the visibility of the interference pattern has minima, and calculate the smallest value on the optical axis, where θ = 0.

Homework Equations

$I=I_0(3+cos(2k_1 x cos\theta )+2cos(2k_2 x cos\theta ))$
minimum of first interference pattern when $4d=n\lambda_1$
minimum of second interference pattern when $4d=m\lambda_2$
The first minimum in intensity comes when $n \lambda_1 = m \lambda_2$

However, i have no idea how to find the lowest values of m and n for which this occurs

The Attempt at a Solution

 

[Charles Link]

This one is very similar to a homework problem a student recently had on the same topic where we did a very thorough treatment of the interference fringe pattern: https://www.physicsforums.com/threa...michelson-interferometer.933638/#post-5902650 $\\$ In the previous homework problem, the spectral lines were assumed to be of equal intensity, but the intensity equations as a function of $\theta$ will hopefully be somewhat helpful. (See post 7 of this thread, etc.) $\\$ One thing the question isn't real clear on here is the "smallest value" at $\theta=0$. Do they want the smallest value of the intensity? The "fringe visibility" normally requires a comparison of the intensity values across the plane, but perhaps they are looking for the smallest value of visibility you can find for fringes near $\theta=0$.[Editing: Upon further study, they do want the smallest value of the visibility for the fringe pattern near $\theta=0$]. $\\$ Incidentally, you never gave any feedback on another recent homework problem you posted. Hopefully you saw my response: https://www.physicsforums.com/threads/radiative-heat-engine.943448/

 

[Physgeek64]

This one is very similar to a homework problem a student recently had on the same topic where we did a very thorough treatment of the interference fringe pattern: https://www.physicsforums.com/threa...michelson-interferometer.933638/#post-5902650 $\\$ In the previous homework problem, the spectral lines were assumed to be of equal intensity, but the intensity equations as a function of $\theta$ will hopefully be somewhat helpful. (See post 7 of this thread, etc.) $\\$ One thing the question isn't real clear on here is the "smallest value" at $\theta=0$. Do they want the smallest value of the intensity? The "fringe visibility" normally requires a comparison of the intensity values across the plane, but perhaps they are looking for the smallest value of visibility you can find for fringes near $\theta=0$. $\\$ Incidentally, you never gave any feedback on another recent homework problem you posted. Hopefully you saw my response: https://www.physicsforums.com/threads/radiative-heat-engine.943448/

So would we say that $n=m+1$
$\delta_1=2k_1 d$
$\delta_2=2k_2 d$
$\delta_1-\delta_2=\pi = 4d(\frac{1}{\lambda_1}-\frac{1}{\lambda_2})$
$d=\frac{\lambda_1 \lambda_2}{4 \Delta \lambda}$

I have a few issues with doing this 1) i don't know why $\delta_1-\delta_2=\pi$ and 2) when i plot this using $\lambda_1=3$ $\lambda_2=3.5$ this value of d does not give the first, deepest minimum, but rather a higher order maximum

Thanks for the reply! :)

 

[Charles Link]

This one is quite a lot simpler than the problem that the previous student had a fringe visibility from a finite spectral line width. $\\$ If you look at post 7 of that link, there is a formula that is derived in post 8 of that link: $\\$ $I(\theta)=I_o \cos^2(\frac{\pi D \cos(\theta)}{\lambda}+\frac{\pi}{2})$, so that $I_{total}(\theta)=I_1 \cos^2(\frac{\pi \, D \, \cos(\theta)}{\lambda_1}+\frac{\pi}{2}) +I_2 \cos^2(\frac{\pi \, D \, \cos(\theta)}{\lambda_2}+\frac{\pi}{2})$. $\\$ Let the two sources be of $\lambda_1= \lambda_o$ and $\lambda_2=\lambda_o+\Delta \lambda$. $\\$ We can write $\frac{\pi \, D \cos(\theta)}{\lambda_o+\Delta \lambda} \approx \frac{\pi D \cos(\theta)}{\lambda_o}-\frac{\pi \, D \, \cos(\theta) \,\Delta \lambda}{\lambda_o^2}$ (with a Taylor expansion of $\frac{1}{\lambda_o+ \Delta \lambda}$). $\\$When $\frac{\pi D \cos(\theta) \, \Delta \lambda}{\lambda_o^2}=m \pi$, (with $\cos(\theta) \approx 1$), the two $\cos^2(A \cos(\theta)+\frac{\pi}{2})$ intensity functions will line up, with maximum basically equal to the total, (three units of intensity) and minimum equal to zero. This will make the visibility $V=\frac{I_{max}-I_{min}}{I_{max}+I_{min}}=1.0$. $\\$ When $\frac{\pi \, D \, \Delta \lambda}{\lambda_o^2}=(2m+1) \frac{\pi}{2}$, the fringe patterns will be complimentary, with $I_{max}$ being only from the 589.0 line, (2 units of intensity), with 589.6 giving zero. What is the $I_{min}$ for this case? Does 589.6 have its maximum at nearly the same $\theta$ that 589.0 has zero intensity? You should be able to compute the visibility for this case. $\\$ (Note in the above, the factor $\frac{\pi D}{\lambda_o}$ is normally quite large, so that small changes in $\cos(\theta)$ can result in many cycles of fringes occurring even as $\cos(\theta)$ for the term $\frac{\pi \, D \, \cos(\theta) \, \Delta \lambda}{\lambda_o^2}$ can be considered to be $\cos(\theta) \approx 1.0$ for most practical purposes. The factor $\frac{\pi \, D \, \Delta \lambda }{\lambda_o^2}=(\frac{\pi \, D }{\lambda_o})(\frac{\Delta \lambda}{\lambda_o})$ is considerably smaller, in this case by a factor of $\frac{1}{1000}$ ).

 

[Charles Link]

For this one, it may be worthwhile to do a quick follow-on: $\\$ Do you see when $D$ is chosen so that the phase term for the $\lambda_1$ source is basically $\pm \frac{\pi}{2}$, (you can ignore the additional $+\frac{\pi}{2}$ phase term that appears in the two intensity terms), that the result is $I_{total}(\theta)=2 \cos^2(\frac{\pi \, D \, \cos(\theta)}{\lambda_o})+\sin^2(\frac{\pi \, D \, \cos(\theta)}{\lambda_o})$? $\\$ You can let $y(x)=2 \cos^2 x +\sin^2 x$ and find the maximum and minimum with standard calculus techniques. You will find for any other phase than $\pm \frac{\pi}{2}$, that $I_{max}>2$ , and $I_{min}<1$.