# Ray transfer matrix - Plano convex lens

• AwesomeTrains
In summary: Nice job!In summary, the ray transfer matrix for a Plano convex lens with a 5 cm radius and refraction index n=1.8 can be found using the paraxial approximation. The matrices for refraction at a curved surface, propagation inside the lens, and refraction at a flat surface can be multiplied to obtain the overall matrix. The direction of the ray must be taken into account when determining the signs of the refraction indices and the radius of curvature.

## Homework Statement

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.

## Homework 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}

## 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 :)

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

@TSny appears to be the expert here.

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:
AwesomeTrains said:
I would multiply the three matrices like this: $\begin{pmatrix} 1 & 0\\ 0 & \frac{n_1}{n_2}\\ \end{pmatrix} \begin{pmatrix} 1 & d\\ 0 & 1\\ \end{pmatrix} \begin{pmatrix} 1 & 0\\ \frac{n_1-n_2}{Rn_2} & \frac{n_1}{n_2}\\ \end{pmatrix}$

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?

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

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

BvU said:
@TSny appears to be the expert here.

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

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)?

AwesomeTrains said:
Refraction at a curved surface:
\begin{pmatrix}
1 & 0\\
\frac{n_1-n_2}{Rn_2} & \frac{n_1}{n_2}\\
\end{pmatrix}
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).

AwesomeTrains said:
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.

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.

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?

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?

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 :)

That looks good to me.

## 1. What is a Ray Transfer Matrix for a Plano Convex Lens?

A Ray Transfer Matrix (RTM) for a Plano Convex Lens is a mathematical tool used to analyze the behavior of light rays passing through a convex lens. It describes the relationship between the incoming and outgoing rays, as well as the lens properties, such as focal length and refractive index.

## 2. How is a Ray Transfer Matrix calculated?

A Ray Transfer Matrix is calculated by multiplying individual matrices that represent the different optical elements in a system, such as a Plano Convex Lens. These matrices take into account the geometry and optical properties of the lens, and can be derived using the laws of geometric optics.

## 3. What are the limitations of using a Ray Transfer Matrix for a Plano Convex Lens?

One limitation of using a Ray Transfer Matrix for a Plano Convex Lens is that it assumes the lens is thin and has a small aperture, which may not hold true for all lenses. Additionally, it only considers paraxial rays and does not account for aberrations or diffraction effects.

## 4. How can a Ray Transfer Matrix be used to design a Plano Convex Lens?

A Ray Transfer Matrix can be used to design a Plano Convex Lens by manipulating the matrix elements to achieve the desired properties, such as focal length and image location. By varying the curvature and refractive index of the lens, the matrix can be optimized to meet specific design criteria.

## 5. What other applications does the Ray Transfer Matrix have besides analyzing Plano Convex Lenses?

The Ray Transfer Matrix has a wide range of applications in optical systems, including analyzing and designing other types of lenses, mirrors, and prisms. It is also used in laser beam propagation, fiber optics, and other optical components and systems.