Would love some clarification with optics

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Hi, I have a major lack of understanding regarding this one issue. The following is consequently pretty incoherent, so I would be even more grateful to anyone who bothers to read and reply.

I can derive the intensity distribution for, say a diffraction grating or single slit, in the Fraunhofer limit. However I really don't understand what happens when a lens is put in the mix.

For example, a diffraction grating with a lens right next to it (not on the same side as the source), observing the pattern in the focal plane: does the intensity distribution have the same angular dependence as before, but with respect to the lens axis?

What about the picture attached (where the focal length is 2.5mm). I can derive, in the Fraunhofer limit, the intensity as a function of angle with respect to the central normal (or perhaps any normal?) to the grating, but have no idea how to get the intensity in the focal plane of the lens (at 7.5mm). How do I find the positions of the maxima and minima? (I'm thinking maybe the central maximum would be on the optical axis, as the normal to the grating is parallel to the axis and so is directed onto the focal point: but how do I know the separation? Is it the same angular dependence as without the lens, but now with respect to the lens axis?).

Sorry for the rambling, any replies would honestly be very much appreciated.
 

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Andy Resnick

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It's tough for me to give a complete answer, but I can give some qualitative information.

The lens will merely move the Fraunhofer (far-field) diffraction pattern to the rear focal plane of the lens. There's a lot of caveats here- the lens law (1/o + 1/i = 1/f) should be obeyed, for example. But essentially, there is no change other than the diffraction pattern being scaled by a factor (wavelength/focal length) that converts the angular dependence (since the Fraunhofer pattern is at infinity) into a linear dependence in the image plane.

The figure you attached is a more complex layout. For example, how is the grating illuminated? Since the grating is not at the front pupil plane, the field at the rear focal plane will pick up some extra phase factors which could greatly alter the diffraction pattern. This may not be relevant, but the image you have reminds me of the Talbot effect:

http://en.wikipedia.org/wiki/Talbot_effect
 
It's tough for me to give a complete answer, but I can give some qualitative information.

The lens will merely move the Fraunhofer (far-field) diffraction pattern to the rear focal plane of the lens. There's a lot of caveats here- the lens law (1/o + 1/i = 1/f) should be obeyed, for example. But essentially, there is no change other than the diffraction pattern being scaled by a factor (wavelength/focal length) that converts the angular dependence (since the Fraunhofer pattern is at infinity) into a linear dependence in the image plane.

The figure you attached is a more complex layout. For example, how is the grating illuminated? Since the grating is not at the front pupil plane, the field at the rear focal plane will pick up some extra phase factors which could greatly alter the diffraction pattern. This may not be relevant, but the image you have reminds me of the Talbot effect:

http://en.wikipedia.org/wiki/Talbot_effect
Thanks for replying.

Maybe the picture isn't a great example. But what about a situation where a single slit is placed at two focal lengths from a lens, and the plane of observation is at the other focal length. I believe this obeys the lens law; so how would one go about finding the intensity distribution on the screen of observation?

Thanks
 

Andy Resnick

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Are you imaging the slit or the transform of the slit? I'm a little confused.
 
Are you imaging the slit or the transform of the slit? I'm a little confused.
Does the transform of the slit mean the diffraction pattern? If so I think I'm talking about that!
 

Redbelly98

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Does the transform of the slit mean the diffraction pattern? If so I think I'm talking about that!
Yes.

The way I picture it is, the angular pattern is produced by sets of parallel "rays" emerging from the slit at each angle. A lens will bring that to a focus (i.e. form an image of the diffraction pattern) at one focal length beyond the lens.

The intensity as a function of angle can be found here:
https://www.physicsforums.com/library.php?do=view_item&itemid=191

With the lens in place, a geometric argument shows that the intensity at angle θ instead appears at a displacement x=θ·f from the central axis (using the usual small-angle approximations).

Hope that helps,

Mark
 
Yes.

The way I picture it is, the angular pattern is produced by sets of parallel "rays" emerging from the slit at each angle. A lens will bring that to a focus (i.e. form an image of the diffraction pattern) at one focal length beyond the lens.

The intensity as a function of angle can be found here:
https://www.physicsforums.com/library.php?do=view_item&itemid=191

With the lens in place, a geometric argument shows that the intensity at angle θ instead appears at a displacement x=θ·f from the central axis (using the usual small-angle approximations).

Hope that helps,

Mark
OK having done a tad more research I think I can finally phrase this properly.

Thanks to your post, I think I get what happens in the focal/Fraunhofer/Fourier plane.

However I still would like to know how to find the intensity distribution in the image plane.

e.g. if I have a grating, which therefore has a square wave amplitude function. I think I would get a square wave function in the image plane, but would the frequency be different to that of the amplitude function at the grating, and if so, how can you work out what it is?

I think I read somewhere that you take a FT of the diffraction pattern in the focal plane, to get the amplitude function in the image plane. Is this correct, and does a geometric argument exist that can be used instead?

Thanks again
 
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rabbit44,

From focal plane to focal plane, the relation is indeed a fourier transform.
However, you still need to find out the scale on which this FT is displayed.
You will find the exact formula in any textbook on Fourier optics.

Then, you are still left with one more problem.
You still need to calculate the steps from the object plane to the input focal plane,
as well as from the output focal plane to the image plane.
 

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