Thick lens problem, transfer matrices

[fluidistic]

Homework Statement

A biconconvex (n_l=1.5) lens have radii worth 20 and 10 cm and an axial width of 5 cm. Describe the image of an object whose height is 2.5 cm and situated at 8 cm from the first vertex.

Homework Equations

Transfer matrices.

The Attempt at a Solution

I used the ray transfer matrix method shown in the website http://physics.tamuk.edu/~suson/html/4323/thick.html.
Let $$M$$ be the ray transfer matrix. Then $$M=M_3M_2M_1$$.
M_1 being a refraction matrix of the first lens, M_2 the ray transfer matrix of the ray passing through the lens and M_3 the refraction of the ray while leaving the lens through the second radius of the lens.

Mathematically $$M_1= \begin{bmatrix}1 & 0 \\ \frac{1}{0.2m} \left ( \frac{1}{1.5}-1 \right ) & \frac{1}{1.5} \end{bmatrix}$$.
$$M_2=\begin{bmatrix}1 & 0.05 \\ 0 & 1 \end{bmatrix}$$.
$$M_3= \begin{bmatrix}1 & 0 \\ -\frac{1}{0.1m} \left ( 1.5-1 \right ) & 1.5 \end{bmatrix}$$.
And thus M gave me $$\begin{bmatrix}1 & \frac{1}{20} \\ 3 & \frac{17}{12} \end{bmatrix}$$.

Therefore $$\begin{bmatrix} y_1 \\ \theta _1 \end{bmatrix}=\begin{bmatrix}1 & \frac{1}{20} \\ 3 & \frac{17}{12} \end{bmatrix} \begin{bmatrix} y_0 \\ \theta _0 \end{bmatrix} =\begin{bmatrix} \frac{1}{40} \\ -\frac{3}{40} \end{bmatrix}$$.
Thus the special ray leaving the maximum height of the object and parallel to the optics axe enter the lens at an height of 1/40 m and leaves the lens at the same height... seems totally counter intuitive to me, therefore wrong.
Anyway it gives me the equation $$y(x)=-\frac{3x}{40}+\frac{1}{20}$$, solving for y(x)=0, I reach a focal distance of 2/3 m! Way too big in my opinion but I'm not 100% sure. Also the image seems real, which totally disagree with my intuition once again since the object is much closer to the lens than the focal length. In disagreement with http://upload.wikimedia.org/wikipedia/commons/9/97/Lens3b.svg, also I know it's a thin lens.
Any help is greatly appreciated.

 

[fluidistic]

I'm still stuck on this exercise. I'd appreciate absolutely any help.

 

[Redbelly98]

Hi fluidistic, I hope you are well.

I haven't used matrix methods in optics very much, but feel that I do understand them reasonably well. I'll see if I can be of help.

Homework Statement

A biconconvex (n_l=1.5) lens have radii worth 20 and 10 cm and an axial width of 5 cm. Describe the image of an object whose height is 2.5 cm and situated at 8 cm from the first vertex.

First, I'll just comment that you could work this out staying in cm units, rather than converting everything to meters. Either way should work in theory, but I think we'd be less prone to making an error if you don't convert the units.

That being said, we may as well go with meters since you have set everything up already.

Homework Equations

Transfer matrices.

The Attempt at a Solution

I used the ray transfer matrix method shown in the website http://physics.tamuk.edu/~suson/html/4323/thick.html.
Let $$M$$ be the ray transfer matrix. Then $$M=M_3M_2M_1$$.
M_1 being a refraction matrix of the first lens, M_2 the ray transfer matrix of the ray passing through the lens and M_3 the refraction of the ray while leaving the lens through the second radius of the lens.

Mathematically $$M_1= \begin{bmatrix}1 & 0 \\ \frac{1}{0.2m} \left ( \frac{1}{1.5}-1 \right ) & \frac{1}{1.5} \end{bmatrix}$$.
$$M_2=\begin{bmatrix}1 & 0.05 \\ 0 & 1 \end{bmatrix}$$.
$$M_3= \begin{bmatrix}1 & 0 \\ -\frac{1}{0.1m} \left ( 1.5-1 \right ) & 1.5 \end{bmatrix}$$.

I agree with your M1, M2, and M3.

And thus M gave me $$\begin{bmatrix}1 & \frac{1}{20} \\ 3 & \frac{17}{12} \end{bmatrix}$$.

I get something different. And my result has a determinant =1, as (I believe) it should be.
Hint: when I multiply M₂M₁, I get

0.917, 0.033m
-1.667m^-1, 0.667

Therefore $$\begin{bmatrix} y_1 \\ \theta _1 \end{bmatrix}=\begin{bmatrix}1 & \frac{1}{20} \\ 3 & \frac{17}{12} \end{bmatrix} \begin{bmatrix} y_0 \\ \theta _0 \end{bmatrix} =\begin{bmatrix} \frac{1}{40} \\ -\frac{3}{40} \end{bmatrix}$$.
Thus the special ray leaving the maximum height of the object and parallel to the optics axe enter the lens at an height of 1/40 m and leaves the lens at the same height... seems totally counter intuitive to me, therefore wrong.

Good observation. You'll need to get the M correct, in order to proceed further.

Anyway it gives me the equation $$y(x)=-\frac{3x}{40}+\frac{1}{20}$$, solving for y(x)=0, I reach a focal distance of 2/3 m! Way too big in my opinion but I'm not 100% sure. Also the image seems real, which totally disagree with my intuition once again since the object is much closer to the lens than the focal length. In disagreement with http://upload.wikimedia.org/wikipedia/commons/9/97/Lens3b.svg, also I know it's a thin lens.
Any help is greatly appreciated.

Okay, I don't follow this last part. Here is what I would try:

I think you'll need to include the free-space travel from the object to the lens, and from the lens to the image, in your overall M for the system. And you'll need to express, mathematically, the fact that all rays from the source converge to the same point at the image location. That is how I would proceed, but if there is a different method that was presented to you in class then I am not sure how else to help.