Ray transfer matrix - Plano convex lens

AwesomeTrains · Dec 29, 2014

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 :)

BvU · Dec 29, 2014

∞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.

AwesomeTrains · Dec 29, 2014

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?

TSny · Dec 29, 2014

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 :)

AwesomeTrains · Dec 30, 2014

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)?

TSny · Dec 30, 2014

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 n_i is the index of refraction on the side of the surface where the light is incident on the surface and n_t 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.

AwesomeTrains · Dec 30, 2014

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?

TSny · Dec 30, 2014

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?

AwesomeTrains · Dec 30, 2014

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 :)

TSny · Dec 30, 2014

That looks good to me.

Ray transfer matrix - Plano convex lens

Homework Statement

Homework Equations

The Attempt at a Solution

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