# Changes

## Holography Transmission Equations Part I

, 17:02, 31 October 2015
m
Introduction
These two papers, plus a follow up in the succeeding 1985 ISDH proceedings by Benton, form the basis for computation of almost all the geometrical problems facing the display holographer. With these papers, the geometry of reference and object angles for pseudo-color holograms (multi-colored replay made with a single color laser) of the transmission and reflection configurations can be plotted. (I have also used them to successfully record HOE’s for a variety of applications.)
Alas and alack, these papers as written are only available in the out of print and hard to find Symposia proceedings. But the equations are available in the sites below, hopefully into perpetuity, as they are the cornerstones of the mathematical foundations of holography, although diagonally opposite.('''Update:''' Courtesy of the author, Stephen McGrew: [http://nli-ltd.com/publications/graphical_method.php ''Graphical Methods''])
Benton’s approach is a trigonometric one, and McGrew’s is a geometric one; they are equivalent, giving the same result, except with Benton you plug and chug through equations, then map out the results. McGrew is more Euclidean; the diagram is made with straightedge and compass.
==Spatial Frequency==
[[Image:InterferenceEQ.gif]]: $\displaystyle f = \frac{sin(\theta_1) - sin(\theta_2)}{\lambda}$
Frequency, how often something happens, the frequencies most often coming to mind are the 20 to 20,000 cycles per second (abbreviated as Hertz) of good human hearing (not mine, for sure!) or the radio frequencies of 550 kiloHertz to 1800 kHZ of the AM band, or the GigaHertz processing speed of computer chips.
Related to this equation is finding the distance between fringes; instead of fringes per mm, how big is a fringe cycle? (= the distance from the center of one bright fringe to the next.) It is simply the inverse of the spatial frequency equation:
[[Image:SpatialEQ.gif]]: $\displaystyle d = \frac{\lambda}{sin(\theta_1) - sin(\theta_2)}$
(And don’t forget theta 2 is a negative angle, otherwise you will have an error!) (Or to make life easier, forget about the sign convention, and just add the sines of the two angles as positives together! But watch your step when getting too far off the track!)
[[Image:DiffractionEQ.gif]]: $\sin{\theta_{out}} = \sin{\theta_{in}} + m \lambda f$
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: $\sin{\theta_{out}} = \sin{(-40)} + 1 (0.00033nm) (1805 c/mm)$:: $\sin{\theta_{out}} = -0.643 + 1.gif]]143$:: $\sin{\theta_{out}} = 0.500$:: $\theta_{out} = 30^\circ$
Just like it should.
If we repeat the above but with m = 2, we get:
[[Image:EdExample2: $\sin{\theta_{out}} = \sin{(-40)} + 2 (0.00033nm) (1805 c/mm)$:: $\sin{\theta_{out}} = -0.643 + 2.286$:: $\sin{\theta_{out}} = 1.gif]]643$
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: $f = \frac{\sin{10} - \sin{(-10)}}{633nm}$:: $f = 548.651...gif]]$
(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: $\sin{\theta_{out}} = m \lambda f$:: $\sin{\theta_{out}} = 1 (.000633nm) (550 c/mm)$:: $\sin{\theta_{out}} = 0.34815$:: $\theta_{out} = 20.gif]]37^\circ$
Replacing m by 2,
[[Image:EdExample5: $\sin{\theta_{out}} = 2 (.000633nm) (550 c/mm)$:: $\sin{\theta_{out}} = 0.6963$:: $\theta_{out} = 44.gif]]1^\circ$
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: $\sin{\theta_{out}} = 1 (.000442nm) (550 c/mm)$:: $\sin{\theta_{out}} = 0.2413$:: $\theta_{out} = 14.gif]]0^\circ$
Replacing m by 2,
[[Image:EdExample7: $\sin{\theta_{out}} = 2 (.gif]]000442nm) (550 c/mm)$:: $\sin{\theta_{out}} = 0.4826$:: $\theta_{out} = 29^\circ$
This time we can get away by replacing m by 3,
[[Image:EdExample8: $\sin{\theta_{out}} = 3 (.000442nm) (550 c/mm)$:: $\sin{\theta_{out}} = 0.gif]]7293$:: $\theta_{out} = 46^\circ$
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)!
1,156
edits