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Holography Transmission Equations Part I

1,097 bytes added, 21:38, 15 August 2015
m
Diffraction
http://courses.media.mit.edu/2002spring/mas450/reading/other_handouts/spatial_freq/index.html
[[Image:DiffractionEQ.gif]]: <math>\sin{\theta_{out}} = \sin{\theta_{in}} + m \lambda f</math>
They plug in the angles, and if you are doing this at home you end up with 1805 lines per mm, which they had rounded to 1800.
Substituting 40 degrees for theta in (a top lit holo in their diagram) and 1805 for f, lambda = 633 nm, same as we made it, that mysterious m set to 1 for right now, we get: (Notice that although I am considering this holo to be top referenced, this goofy MIT sign convention considers it a negative angle!)
[[Image:EdExample1: <math>\sin{\theta_{out}} = \sin{(-40)} + 1 (0.00033nm) (1805 c/mm)</math>:: <math>\sin{\theta_{out}} = -0.643 + 1.gif]]143</math>:: <math>\sin{\theta_{out}} = 0.500</math>:: <math>\theta_{out} = 30^\circ</math>
Just like it should.
If we repeat the above but with m = 2, we get:
[[Image:EdExample2: <math>\sin{\theta_{out}} = \sin{(-40)} + 2 (0.00033nm) (1805 c/mm)</math>:: <math>\sin{\theta_{out}} = -0.643 + 2.286</math>:: <math>\sin{\theta_{out}} = 1.gif]]643</math>
Unh-unh! sin of an angle can’t be >1! (This is called a [[http://en.wikipedia.org/wiki/Evanescent_wave evanescent]] wave and is not propegatedpropagated.)
So what is the role of m? Let’s make a grating with a lower spatial frequency. (an unslanted grating with beams at a small angle to the plate)
[[Image:EdExample3: <math>f = \frac{\sin{10} - \sin{(-10)}}{633nm}</math>:: <math>f = 548.651...gif]]</math>
(Damn, why can’t you cut and paste numbers from the Windows calculator into a document! And why does it disappear when I start typing!) Anyhow, let’s just round our spatial frequency to 550 cycles per mm.
Let’s do something different when interrogating this grating this time: after processing for ultimate efficiency, let’s hit the grating with a raw, undiverged He-Ne beam along the normal, so that way sin theta in disappears (sin 0 = 0) and our output equation becomes:
[[Image:EdExample4: <math>\sin{\theta_{out}} = m \lambda f</math>:: <math>\sin{\theta_{out}} = 1 (.000633nm) (550 c/mm)</math>:: <math>\sin{\theta_{out}} = 0.34815</math>:: <math>\theta_{out} = 20.gif]]37^\circ</math>
Replacing m by 2,
[[Image:EdExample5: <math>\sin{\theta_{out}} = 2 (.000633nm) (550 c/mm)</math>:: <math>\sin{\theta_{out}} = 0.6963</math>:: <math>\theta_{out} = 44.gif]]1^\circ</math>
When m = 3, we get a sine > 1, so we’re out of luck. But we could also replace m by -1 and -2, getting -20.4 and -44.1 degrees as more output angles.
To see where the other end of the rainbow comes out, let’s replace lambda by another popular Helium atmosphere laser, Helium-Cadmium, at 442 nm
[[Image:EdExample6: <math>\sin{\theta_{out}} = 1 (.000442nm) (550 c/mm)</math>:: <math>\sin{\theta_{out}} = 0.2413</math>:: <math>\theta_{out} = 14.gif]]0^\circ</math>
Replacing m by 2,
[[Image:EdExample7: <math>\sin{\theta_{out}} = 2 (.gif]]000442nm) (550 c/mm)</math>:: <math>\sin{\theta_{out}} = 0.4826</math>:: <math>\theta_{out} = 29^\circ</math>
This time we can get away by replacing m by 3,
[[Image:EdExample8: <math>\sin{\theta_{out}} = 3 (.000442nm) (550 c/mm)</math>:: <math>\sin{\theta_{out}} = 0.gif]]7293</math>:: <math>\theta_{out} = 46^\circ</math>
So if we hit this grating with the He-Cd, dead nuts on at a right angle, we would see 7 beams: the straight through zero order, (see what happens when m = 0 in the above), two beams on either side of the normal at 14, 29 and 46 degrees (corresponding to m=-3,-2,-1,0,1,2 and 3)!

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