# Ray transfer matrix - Plano convex lens

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1. Dec 29, 2014

### AwesomeTrains

1. The problem statement, all variables and given/known data
I have to find the ray transfer matrix for a Plano convex lens, with a 5 cm radius and refraction index n=1.8.
I'm supposed to use the paraxial approximation.
Attached a homemade picture of the lens.

2. Relevant equations
Refraction at a curved surface:
\begin{pmatrix}
1 & 0\\
\frac{n_1-n_2}{Rn_2} & \frac{n_1}{n_2}\\
\end{pmatrix}
Propagation matrix for the inside:
\begin{pmatrix}
1 & d\\
0 & 1\\
\end{pmatrix}
Refraction at a flat surface:
\begin{pmatrix}
1 & 0\\
0 & \frac{n_1}{n_2}\\
\end{pmatrix}
3. The attempt at a solution
I would multiply the three matrices like this: $\begin{pmatrix} 1 & 0\\ 0 & \frac{n_1}{n_2}\\ \end{pmatrix} \begin{pmatrix} 1 & d\\ 0 & 1\\ \end{pmatrix} \begin{pmatrix} 1 & 0\\ \frac{n_1-n_2}{Rn_2} & \frac{n_1}{n_2}\\ \end{pmatrix}$

What is d? Is it 5 cm because it's a half sphere and all rays travel 5 cm?

Kind regards Alex :)

2. Dec 29, 2014

### BvU

∞Looks correct to me. "center thickness of lens" is d allright and R2 = ∞. You might want to check this thread.

@TSny appears to be the expert here.

3. Dec 29, 2014

### AwesomeTrains

Great thanks for the help but the result depends on the direction from which the ray comes. If it hits the flat surface I get the same result as the link you sent me but the other way, I get some other result: $\frac{n_2R}{n_2-n_1}$ with n_1=1. What result is correct?

Last edited: Dec 29, 2014
4. Dec 29, 2014

### TSny

Something to think about: Do n1 and n2 in the first matrix have the same values as n1 and n2 in the last matrix? In other words, what is the interpretation of n1 and n2 at a refracting surface?

Yes, for paraxial rays, d can be approximated by the thickness of the lens.

A very kind remark. But, sadly, I know very little optics. However, I do have an old copy of Hecht and Zajac :)

5. Dec 30, 2014

### AwesomeTrains

Thanks for the replies.
Yea TSny that was my first mistake :) Corrected it:

$\begin{pmatrix} 1 & 0\\ 0 & \frac{n_1}{n_2}\\ \end{pmatrix} \begin{pmatrix} 1 & d\\ 0 & 1\\ \end{pmatrix} \begin{pmatrix} 1 & 0\\ \frac{n_2-n_1}{Rn_1} & \frac{n_2}{n_1}\\ \end{pmatrix}$
It is reversed in the last matrix because the ray goes from lens to air.

Then I get this:
$\begin{pmatrix} 1+\frac{d(n_2-n_1)}{Rn_1} & \frac{dn_2}{n_1}\\ \frac{n_2-n_1}{Rn_2} & 1\\ \end{pmatrix} then \frac{n_2-n_1}{Rn_2}=\frac{1}{f}$

If I do it the otherway around I get the other result:

$\begin{pmatrix} 1 & \frac{dn_1}{n_2}\\ \frac{n_2-n_1}{Rn_1} & \frac{d(n_2-n_1)}{Rn_2}+1\\ \end{pmatrix} then \frac{1}{f}=\frac{n_2-n_1}{Rn_1}$
Which result is correct when you consider the object is on the left side, facing the curved surface. (First matrix)?

6. Dec 30, 2014

### TSny

In order to get the n1 and n2 in the right place, I find it useful to write the matrix as

\begin{pmatrix}
1 & 0\\
\frac{n_i-n_t}{Rn_t} & \frac{n_i}{n_t}\\
\end{pmatrix}
where ni is the index of refraction on the side of the surface where the light is incident on the surface and nt is the index on the other side of the surface (the transmitted side).

The last matrix is the refraction matrix for the first surface, where the light goes from air into the lens. The last matrix will act first when operating on a ray. You have to get used to reading matrix products from right to left when they operate on a ray.

The overall matrix for the case of light hitting the flat surface first will not be the same as the case of light hitting the curved surface first, but you should get the same focal length for both cases.

Let's check our matrices again, I'm not getting the same result as you.

7. Dec 30, 2014

### AwesomeTrains

Thanks for sticking with me :) I did the calculation again:
For a ray incident on the curved surface:
$\begin{pmatrix} 1 & 0\\ 0 & \frac{n_2}{n_1}\\ \end{pmatrix} \begin{pmatrix} 1 & d\\ 0 & 1\\ \end{pmatrix} \begin{pmatrix} 1 & 0\\ \frac{n_1-n_2}{Rn_2} & \frac{n_1}{n_2}\\ \end{pmatrix} = \begin{pmatrix} 1+\frac{d(n_1-n_2)}{Rn_2} & \frac{dn_1}{n_2}\\ \frac{n_1-n_2}{Rn_1} & 1\\ \end{pmatrix}$
And for a ray incident on the flat surface:
$\begin{pmatrix} 1 & 0\\ \frac{n_2-n_1}{Rn_1} & \frac{n_2}{n_1}\\ \end{pmatrix} \begin{pmatrix} 1 & d\\ 0 & 1\\ \end{pmatrix} \begin{pmatrix} 1 & 0\\ 0 & \frac{n_1}{n_2}\\ \end{pmatrix} = \begin{pmatrix} 1 & \frac{dn_1}{n_2}\\ \frac{n_2-n_1}{Rn_1} & \frac{d(n_2-n_1)}{Rn_2}+1\\ \end{pmatrix}$
The C entries are slightly different again though, have I messed up the refraction indices again?

8. Dec 30, 2014

### TSny

OK. That's very close to what I get. For the second case where the light is incident on the flat surface, did you account for the sign of the radius of curvature of the curved surface?

9. Dec 30, 2014

### AwesomeTrains

Ah thanks for the hint, that explains the sign difference :)
Corrected for the flat surface:
$\begin{pmatrix} 1 & 0\\ \frac{n_2-n_1}{(-R)n_1} & \frac{n_2}{n_1}\\ \end{pmatrix} \begin{pmatrix} 1 & d\\ 0 & 1\\ \end{pmatrix} \begin{pmatrix} 1 & 0\\ 0 & \frac{n_1}{n_2}\\ \end{pmatrix} = \begin{pmatrix} 1 & \frac{dn_1}{n_2}\\ \frac{n_2-n_1}{(-R)n_1} & \frac{d(n_2-n_1)}{(-R)n_2}+1\\ \end{pmatrix}$
Now it fits, thanks a lot :)

10. Dec 30, 2014

### TSny

That looks good to me.