# Microscope magnification using Ray Optics

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1. May 11, 2015

### sam400

1. The problem statement, all variables and given/known data

Basically, derive the formula $m = \frac{ 25 cm}{f_e} \frac{L}{f_o}$ using ray matrices. This just has variable tube length and assumes eye to object distance is 25 cm.

2. Relevant equations

Ray matrices: $\left[ \begin{array}{cc} 1 & d \\ 0 & 1 \end{array} \right]$
$\left[ \begin{array}{cc} 1 & 0 \\ - \frac{1}{f} & 1 \end{array} \right]$
ray vector: $\left[ \begin{array}{c} r \\ \theta \end{array} \right]$

$M = \frac{ \theta '}{\theta}$

3. The attempt at a solution

So, just set up the ray transform
$\left[ \begin{array}{cc} 1 & 25 \\ 0 & 1 \end{array} \right] \left[ \begin{array}{cc} 1 & 0 \\ a & 1 \end{array} \right] \left[ \begin{array}{cc} 1 & b \\ 0 & 1 \end{array} \right] \left[ \begin{array}{cc} 1 & 0 \\ c & 1 \end{array} \right] \left[ \begin{array}{c} r \\ \theta \end{array} \right]$

Here, $a = - \frac{1}{f_e}, b = f_o + f_e + L, c = - \frac{1}{f_o}$

$f_e$ is eyepiece focal length and $f_o$ is objective one. Eitherway, I just simplified the matrix given above, but I can't seem to make it the same as the formula given. I had another matrix at first with some distance $d$ but that just made it messier. I'm not sure if I need that. But for the $\theta '$ component, I get

$\theta ' = \frac{ L r} { f_e f_o}$
the other term with $\theta$ should just be 0 since the rays will not depend on the initial angle and all will be parallel. I know the initial problem says the object is 25 cm away, so does that mean r = 25 cm? But that would still leave the problem of dividing the $\theta$ so not sure where to go.