# Homework Help: Optics Matrix Methods

1. Nov 26, 2006

### Walczyk

So I've gotten through half of this very long problem involving thick lens matrices and using the calculated system matrix for image locations. I need help at part d!!!! I'm going mad with it.

$$S = \begin{array}{cc}\frac{11}{8}&\frac{5}{2}\\\frac{3}{20}&1\end{array}$$

So I ended up with a system matrix with elements a11= 11/8 a12= 5/2 a21= 3/20 a22= 1. Calculations were S1=

r2 x t1 x r1. Looks good so far.

I did this correct I think, I'm trying to get my scanner to work now.
I got an answer of -60/11 cm, that's to the left of the right side of the lens.

I got a system matrix of elements a11= 11/8 + 6x/40, a12= 11s/8 + 6sx/40 + x + 5/2, a21= 6/40, a22= 1 +

6s/40. I took the original three refraction and transfer matrices and added two more transfer matrices, one

for s and x. Calculations were S1= t3 x r2 x t2 x r1 x t1.

I have no idea how to do this. I figured out the focal lengths, and they came out to be the same as in part

B, but the matrix in part D has the first two elements with respect to x when s= 30cm. Do I use matrix

algebra to solve for x? What about magnification. I think I missed a lecture.

The rest of the problem:
It acts like a thin lens doesn't it? p and q will be zero, focal points will be on their respective input and

output planes. That's all I have figured out.

Simple; you multiply the system matrix by an object matrix o11, o21 to solve an image matrix i11, i21 with

height and angle as elements.

Last edited: Nov 27, 2006