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