# Changes

## Holography Transmission Equations Part I

, 16:43, 15 May 2013
m
Reverted edits by Jsfisher (talk) to last revision by Colin Kaminski
==Spatial Frequency==
[[Image: $f = \frac{\sin{\theta_2} - \sin{\theta_1}}{\lambda}$InterferenceEQ.gif]]
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: $d = \frac{\lambda}{\sin{\theta_2} - \sin{\theta_1}}$SpatialEQ.gif]]
(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!)
Alternatively, to find the fringe spacing distance, you could simply use the 1/x button on the calculator once you have your spatial frequency. In either case the distance will be in millimeters, and the digits will usually all pop up behind the decimal point, so if you want to convert to microns, move the decimal point over three places to the right.
Onward and upward, as my Abstract Algebra teacher used to say . (while I nodded off!).
[[Image: $\sin{\theta_{out}} = \sin{\theta_{in}} + m \lambda f$DiffractionEQ.gif]]
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: $\begin{array}{rcl}\sin{\theta_{out}} & = & \sin{(-40)} + 1 \times 0EdExample1.000633 \, mm \times 1805 \, c/mm \\\sin{\theta_{out}} & = & -0.643 + 1.143 \\\sin{\theta_{out}} & = & 0.5 \\\theta_{out} & = & 30^\circ\end{array}$gif]]
Just like it should.
If we repeat the above but with m = 2, we get:
[[Image:EdExample2.gif]]
: $\begin{array}{rcl}\sin{\theta_{out}} & = & \sin{(-40)} + 2 \times 0.000633 \, mm \times 1805 \, c/mm \\\sin{\theta_{out}} & = & -0.643 + 2.286 \\\sin{\theta_{out}} & = & 1.643\end{array}$  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 propogatedpropegated.)
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.gif]]
: $\begin{array}{rcl} f & = & (\sin{(10)} - \sin{(-10)}) / 633 \, nm \\ f & = & 548.651... \end{array}$
(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: $\begin{array}{rcl}\sin{\theta_{out}} & = & m \lambda f \\\sin{\theta_{out}} & = & 1 \times 0EdExample4.000633 \, mm \times 550 \, c/mm \\\sin{\theta_{out}} & = & 0.34815 \\\theta_{out} & = & 20.37^\circ\end{array}$gif]]
Replacing m by 2,
[[Image: $\begin{array}{rcl}\sin{\theta_{out}} & = & 2 \times 0EdExample5.000633 \, mm \times 550 \, c/mm \\\sin{\theta_{out}} & = & 0.6963 \\\theta_{out} & = & 44.1^\circ\end{array}$gif]]
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

: $\begin{array}{rcl} \sin{\theta_{out}} & = & 1 \times 0.000442 \, mm \times 550 \, c/mm \\ \sin{\theta_{out}} & = & 0.2413 \\ \theta_{out} & = & 14.0^\circ \end{array}$
[[Image:EdExample6.gif]]
Replacing m by 2,

: $\begin{array}{rcl} \sin{\theta_{out}} & = & 2 \times 0.000442 \, mm \times 550 \, c/mm \\ \sin{\theta_{out}} & = & 0.4826 \\ \theta_{out} & = & 29^\circ \end{array}$
[[Image:EdExample7.gif]]
This time we can get away by replacing m by 3,

: $\begin{array}{rcl} \sin{\theta_{out}} & = & 3 \times 0.000442 \, mm \times 550 \, c/mm \\ \sin{\theta_{out}} & = & 0.7293 \\ \theta_{out} & = & 46^\circ \end{array}$
[[Image:EdExample8.gif]]
==Angle Equation==

: $\sin{\theta_{out}} = m \frac{\lambda_{ill}}{\lambda_{exp}}(\sin{\theta_{obj}} - \sin{\theta_{ref}}) + \sin{\theta_{ill}}$
[[Image:AngleEq.gif]]
Preserving the recording parameters (putting the holo back in its plateholder and lighting it up under the exposure conditions):

: $\begin{array}{rcl} \sin{\theta_{out}} & = & m (\lambda_{ill}/\lambda_{exp})(\sin{\theta_{obj}} - \sin{\theta_{ref}} + \sin{\theta_{ill}} \\ \sin{\theta_{out}} & = & 1 (633 \, nm / 633 \, nm)(\sin{(0)} - \sin{(-45)} + \sin{(-45)} \\ \sin{\theta_{out}} & = & 1 \times 1 \times 0.7071 - 0.7071 \\ \sin{\theta_{out}} & = & 0 \\ \theta_{out} & = & 0^\circ \end{array}$
[[Image:AngleEqEx1.gif]]
Moving the reference beam 5 degrees upward to 50 degrees:

: $\begin{array}{rcl} \sin{\theta_{out}} & = & 1 (633 \, nm / 633 \, nm)(\sin{(0)} - \sin{(-45)} + \sin{(-50)} \\ \sin{\theta_{out}} & = & 1 \times 1 \times 0.7071 - 0.7660 \\ \sin{\theta_{out}} & = & -0.0589 \\ \theta_{out} & = & 3.337^\circ \end{array}$
[[Image:AngleEqEx2.gif]]
Moving the reference beam 5 degrees downward to 40 degrees:

: $\begin{array}{rcl} \sin{\theta_{out}} & = & 1 (633 \, nm / 633 \, nm)(\sin{(0)} - \sin{(-45)} + \sin{(-40)} \\ \sin{\theta_{out}} & = & 1 \times 1 \times 0.7071 - 0.6428 \\ \sin{\theta_{out}} & = & 0.0643 \\ \theta_{out} & = & 3.687^\circ \end{array}$
[[Image:AngleEqEx3.gif]]
1,156
edits