# Changes

## Holography Transmission Equations Part I

, 22:17, 14 May 2013
m
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==Spatial Frequency==
<center>: $\displaystyle f = \frac{\sin{\theta_2} - \sin{\theta_1}}{\lambda}$</center>
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:
<center>: $d = \frac{\lambda}{\sin{\theta_2} - \sin{\theta_1}}$</center>
(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!)
: $\begin{array}{rcl} f & = & \frac{(\sin{(10)} - \sin{(-10)}}{) / 633 \, nm} \\ f & = & 548.651... \end{array} : [itex]\begin{array}{rcl}\sin{\theta_{out}} & = & m \lambda f \\\sin{\theta_{out}} & = & 1 \times 0.000633 \, mm \times 550 \, c/mm \\\sin{\theta_{out}} & = & 0.34815\\ \theta_{out} & = & 20.37^\circ \end{array}$
: $\begin{array}{rcl}\sin{\theta_{out}} & = & 2 \times 0.000633 \, mm \times 550 \, c/mm \\\sin{\theta_{out}} & = & 0.6963\\ \theta_{out} & = & 44.1^\circ \end{array}$
: $\begin{array}{rcl}\sin{\theta_{out}} & = & 1 \times 0.000442 \, mm \times 550 \, c/mm \\\sin{\theta_{out}} & = & 0.2413\\ \theta_{out} & = & 14.0^\circ \end{array}$
[[Image:EdExample6.gif]]
: $\begin{array}{rcl}\sin{\theta_{out}} & = & 2 \times 0.000442 \, mm \times 550 \, c/mm \\\sin{\theta_{out}} & = & 0.4826\\ \theta_{out} & = & 29^\circ \end{array}$
[[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]]
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