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## Homework Statement

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.

## Homework 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} ##

## 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.

Thanks in advance.