How Can Thick Lens Matrix Methods Locate Image Positions?

[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.

A thick glass lens (n=1.6) whose first surface has a radius of curvature of +5cm and a

thickness of 4cm and whose second surface is flat, is in air.

a) Starting with refraction and translation matrices, write down and simplify the system matrix for the lens

alone, taking the input and output planes to be at the start and end of the lens respectively. Please use the

"in class" method.

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

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.

b) Using the matrix elements from part a, locate all cardinal points and sketch them on a scale

diagram, showing and labeling all distances.

I did this correct I think, I'm trying to get my scanner to work now.

c) Using the object, image focal length equation (-f1/So + f2/Si = 1) locate the image of an object placed 30

cm in front of the lens. Be specific and give the location relative to the flat side of the lens.

I got an answer of -60/11 cm, that's to the left of the right side of the lens.

d) Now consider an object located at a distance s in front of the lens and its image, located at x

beyond the lens. Now write down and simplify the extended system matrix, taking the input plane to be at s

and the output plane to be at x. Keep s and x as variables in your answer.

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.

e) Using the matrix elements from your matrix in part d), find the image location and the

magnification if your object is placed at s= 30cm. Does your result match your calculation in part c? I hope

so!

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:

f) What information can you get from setting matrix elements A and D equal to zero in your matrix from

part d)? Try it. Do your results agree with your diagram from part b)? Explain.

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.

g) For this situation in part e), trace a ray (mathematically) emanating from a point on the object

0.75cm above the optical axis at an angle of (2/3) radians too determine the height and angle of the ray at

the image location.

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.

 

[anathema]

Thank you for sharing your progress and questions regarding the thick lens matrices problem. I understand that you are currently stuck on part d and are looking for some guidance. I will do my best to explain and help you through this part.

Firstly, I want to address your concern about using matrix algebra to solve for x. Yes, that is the correct approach. You can solve for x by setting the matrix elements A and D equal to zero and then using matrix algebra to solve for x. This will give you the location of the image relative to the flat side of the lens.

As for the magnification, you can use the formula M = -Si/So, where Si is the image distance and So is the object distance. You have already calculated the image distance in part d, so you can substitute that value into the formula to find the magnification.

In terms of the rest of the problem, you are correct in your understanding that setting matrix elements A and D equal to zero will result in a thin lens situation. This is because in a thin lens, the focal points are on the input and output planes, as you mentioned.

For part g, you can use the same approach as in part c to find the image height and angle. You can use the formula h' = Mh, where h' is the image height, M is the magnification, and h is the object height. The angle of the ray can be found using the formula tan(theta) = h'/x, where theta is the angle of the ray, h' is the image height, and x is the distance from the lens to the image.

I hope this helps you understand and solve part d and the rest of the problem. If you have any further questions, please do not hesitate to ask. Keep up the good work and don't get discouraged, you are making great progress!

 

[kyle8921]

Then you multiply the image matrix by the system matrix and solve for the

final image matrix with height and angle as elements. This will give you the height and angle of the ray at

the image location.

Dear student,

It seems like you have made good progress on this problem involving thick lens matrices and the use of system matrices for image locations. However, it is understandable that you are feeling overwhelmed and need help with part d. Let me try to provide some guidance and clarification.

In part d, you are asked to consider an object located at a distance s in front of the lens, and its image located at a distance x beyond the lens. The goal is to write down and simplify an extended system matrix that takes into account the input plane at s and the output plane at x. The key here is to remember that the system matrix represents the transformation of rays from the input plane to the output plane. So, the extended system matrix should include the transformations at both the input plane (s) and the output plane (x).

To do this, you can start with the original three refraction and transfer matrices, as you mentioned, and then add two more transfer matrices for s and x. The transfer matrix for s will have elements that correspond to the distance s, and the transfer matrix for x will have elements that correspond to the distance x. You can then multiply all five matrices together in the correct order to get the final extended system matrix.

Once you have the extended system matrix, you can use it to find the image location and magnification for an object placed at s=30cm. To do this, you can use the equation for image location (-f1/So + f2/Si = 1) and the equation for magnification (m = -Si/So) and plug in the appropriate elements from the extended system matrix. Your result should match the calculation in part c, as the extended system matrix takes into account the distance s and x, which were not considered in part c.

In part f, you are asked to set matrix elements A and D equal to zero in the extended system matrix from part d and see what information you can get. This essentially means setting the focal length of the lens to zero, which makes it act like a thin lens. As you mentioned, this will make the focal points on their respective input and output planes, and the focal lengths will be zero. This will result in a thin lens system, which is easier