Register to reply 
Grating Spacing With Fourier Optics 
Share this thread: 
#1
Jan2413, 11:31 PM

P: 5

Just ahead of time, no this is not related to homework or coursework in my case. I am a TA for an optics lab and need to know it so I can help the students in my class.
I am trying to find an expression for the xspacing for a grating imaged using the 4f method, which is a grating that is 1 focal length away from the first lens, which is then 1f away from the fourier plane, which is 1f away from the second lens which has a focal length equal to the first, which is finally 1f away from the image plane. Specifically, I am interested in knowing what the spacing is for the Fourier transform of a grating when it passes through a Fourier lens. I've heard it is somehow proportional to f*lambda/b, where b is the spacing between slits in the grating, f is the focal length of the lens and lambda is the wavelength of the laser, but I have not been able to find a source that solves such a problem. I would appreciate it if you could give me either an equation for the spacing or a more direct method for solving it. So far, the only methods I've found end with a kspace solution and don't try to translate the peaks you would see to actual displacements (xcoordinate). 


#2
Jan2513, 08:33 AM

Sci Advisor
P: 5,513

You haven't mentioned if it's an amplitude or a phase grating. For an amplitude grating (Ronchi ruling), the transmission function can be written as T = rect(bx)**comb(ax), where 'b' is the width of a single "slit" and 'a' the spacing between slits. '**' means convolution, not multiplication, so the transmission function is interpreted as a row of slits of width 'b' with a centertocenter spacing of 'a'. If this is illuminated with a plane wave, the FT is sinc(x'/b)*comb(x'/a), where '*' is multiplication and x' = x/(lambda*f) due to the focal length of the lens and illumination wavelength. This demonstrates that decreasing the slit width and/or spacing broadens the diffraction pattern, and illumination with decreasing wavelength and/or focal length also broadens the pattern.
For a phase grating, the transmission function is now exp(iT) (if the phase grating is a squarewave type, otherwise substitute in your own phase function) and the FT is significantly more complicated and qualitatively different than the amplitude case. Even so, decreasing the width and/or spacing of the phase grating broadens the pattern, etc. Detailed derivations are easily found: http://books.google.com/books?id=ow5...nsform&f=false http://ocw.mit.edu/courses/mechanica...1S09_lec16.pdf 


#3
Jan2613, 01:02 AM

P: 5

This is an amplitude grating. So just to clarify then, the FT would look, in the spatial domain, like a series of spikes with a sinc function describing their amplitudes from a defined center? Also, the spacing between said spikes would be lambda*f/a?



Register to reply 
Related Discussions  
How do I find the spacing(d) of a grating in a diffraction grating experiment.  Introductory Physics Homework  1  
Spacing of diffraction grating  Introductory Physics Homework  1  
Measuring spacing of diffraction grating with spectroscope  Introductory Physics Homework  0  
The Diffraction Grating  Fringe Brightness and Spacing  Introductory Physics Homework  0  
Order of magnitude of Grating Spacing!  Advanced Physics Homework  1 