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Fabry-Perot interferometer diffraction

  • Thread starter Math Jeans
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  • #1
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Homework Statement


A Fabry-Perot interferometer consists of two parallel half-silvered mirrors separated by a small distance a. Show that when light is incident on the interferometer with an angle of incidence [tex]\theta[/tex], the transmitted light will have maximum intensity when [tex]a = (\frac{m \cdot \lambda}{2})cos(\theta) [/tex]


Homework Equations


[tex]d \cdot sin(\theta) = m \cdot \lambda[/tex]


The Attempt at a Solution



I tried relating the distance a with d by creating a right triangle from the reflected rays.

That method didn't work, and my answer was the same as the one that I was trying to prove, but divided by (sin([tex]\theta[/tex]))^2. I need a new way to do this.
 
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Answers and Replies

  • #2
alphysicist
Homework Helper
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Hi Math Jeans,

The equation [itex]d \sin\theta=m\lambda[/itex] is a derived equation that applies to specific cases (e.g., two slit interference experiment).

The more fundamental relationship is:

[tex]
\mbox{(path length difference of two rays)}=m\lambda
[/tex]

gives constructive interference. (Assuming coherent light that starts out in phase, and assuming that the path length difference effect is the only interference effect).

So when light strikes one of the mirrors inside the interferometer, some passes through the mirror and some is reflected. The maximum will occur when the extra distance that the reflected light has to travel before reaching that same mirror is equal to [itex]m\lambda[/itex].

What does using the right triangles indicate that the extra distance is? Once you have that you can find the given answer. What do you get?
 
  • #3
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The reason that I used dsin(theta) as my phase difference, is because the light entering the interfermometer exits in such a way that there is a distance between the exiting rays that when applied to dsin(theta) will still exibit the same phase difference. It should be working, but it doesn't seem to be.

http://hyperphysics.phy-astr.gsu.edu/Hbase/phyopt/fabry.html" [Broken]

That link has info the the interfermometer, and a diagram (the second one down the page) that shows light entering and exiting it. What I was doing in my calculations was equating the distance between the exiting rays to d, and using the rays bouncing back in forth to form my right triangle. The dsin(theta) relation should work for this.
 
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  • #4
alphysicist
Homework Helper
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Math Jeans,

I think you might be focusing on the wrong distance in this problem. When we have a diffraction grating or a two slit source the light rays are in phase (at least when we use [itex]d\sin\theta[/itex]). If that is the case, the important distance would be the distance between exit points along the mirror. But here, the phase of the light rays at the exit points are not automatically in phase; the point is we have to choose how the separation a and angle theta are related to make them be in phase when the rays leave the mirror.


So this Fabry-Perot problem is somewhat more like a thin-film type of problem than like a two-source interference problem.

When part of the light ray exits the mirror, the part that remains has to reflect back to the other mirror, then reflect again. The distance the light travels between mirrors is [itex]d=a/\cos\theta[/itex], and the light travels twice this distance between adjacent exit points.

I've created a quick picture at:

http://img215.imageshack.us/my.php?image=fpjl4.jpg

The extra distance that one light ray has to travel more than the one that previously exited is the distance in color. (The color was red, but seemed to change after uploading.)

If I misunderstood your post please let me know.
 
  • #5
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Ohhhh. I think that I understand now. I was focusing only on the distance between the exit points as the phase difference of the light, when I should have been focusing on the distance that the light traveled between exit points.

I appreciate your help.
 

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