Application of matrix method to thick lens

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Optics series lecture, lecture — VI.

All Optics series articles here

Application of matrix method to thick lenses

This lecture was delivered on 7th February 2017, in a lecture session of 1 and 1/2 hours. This lecture is slightly bulged out so it might take a little more time than intended.

Most of the lectures are intended for honors courses but once in a while the optics series contains lectures suitable for elective courses. All lectures of this series can be accessed at the link just provided. 


Photo-credit: click on link to left.

Topics covered in this lecture 

A. Cardinal points 

B. Thick lens equation and matrix for thick lens 

C. System matrix for thin lens 

D. Unit and Nodal planes 

E. Matrix for a system of 2 thin lenses


There is a previous article on matrix methods for optical systems, here. It introduces the method. I recommend you read that article as well, preferably before the current one. 

Our previous studies of optical systems were based on two premises.

a. We assumed a paraxial system

b. We assumed that our lenses are thin

a. We assumed a paraxial system, see what this means and how its defined in a previous article, here.

This means we employed a first order optical theory. Check the article just linked for a good overview of whats paraxial optics and whats first order optical theory. Such assumptions are fraught with various types of aberrations, which we studied in detail in lectures, here and here.

b. We assumed that our lenses are thin

This we did for simplicity. In physics when we assume a simple situation, we are not evading the actual complexity of the situation, we are just postponing this to the happy hour, howsoever one defines it.

Some people go by the Friday happy hour rule. It gives a good substratum on which a disposition can be carried out. Later one develops the nuances and fits it into the substratum and if things are carried out with caution and skill one gets a very effective overview of the pedagogy.

Let us now delve into the complexity of the optical system, as a next step from its simple substratum of a thin lens. Our analysis needs to be modified for applying optical principles to optical systems when we consider thick lenses.

In our last lecture, here, lecture — V, we studied the method of matrices in understanding optical ray tracing. Let us now apply this method to the case of thick lens and see what power it unleashes.

Thick lens

I. Cardinal points

A thick lens can be considered to be an equivalent of multiple thin lenses.

Let us draw two suitable thick lens diagrams.


Let’s first discuss the parameters of the thick lens system.

First focal point or object focus, F_o; Rays emerging from F_o, incident on the first spherical surface, will emerge parallel to optical axis. Check the first diagram above.

Second focal point or image focus, F_i; If rays incident on first spherical surface are parallel, then they meet at F_i after refraction from second spherical surface. Check the 2nd diagram above.

The forward focal length, F.F.L.; The forward focal length is the distance between object focus F_o and vertex to the first spherical surface, V_1.

The backward focal length, B.F.L.; The backward focal length is the distance between image focus F_i and vertex to second spherical surface, V_2.

Image surfaces; When rays from F_o and F_i object and image focuses, are extrapolated they meet on a curved surface or line, rather than a plane or straight line. Such a locus of points can lie outside of the lens. The surface coincides with the principal planes in paraxial region, — see principal planes below.

The principal planes; These planes are tangent to the surface which is created when rays from F_o and F_i are extrapolated. The principal planes meet the optical axis at 1st and 2nd principal points H_1 and H_2.

Nodal points; Any ray going through optical center emerges parallel to incident ray — see the diagram below. When incident ray and emergent rays are extrapolated they meet at points N_1 and N_2 on optical axis. N_1 is called as first nodal point and N_2 is called as second nodal point.

When lens is surrounded by same medium on both sides — and of-course has a symmetric geometry, then the nodal and the principal points are the same.

The six points that we just discussed, F_o and F_i, principal points H_1 and H_2 and nodal points N_1 and N_2, are together known as cardinal points of the system.

The length between H_1 and H_2 is nearly equal to \frac{1}{3}\,rd of the length between V_1 and V_2 for ordinary glass lenses in air.


II. Thick lens equations

Thick lenses are equivalent to two spherical refracting surfaces separated by a distance d_1 between the vertices. We can write a relation between image distance v, object distance u and the focal length f  , for a thick lens as follows, given that; object distance u   is measured from first principal plane H_1   and image distance v is measured from second principal plane H_2:

\boxed{\frac{1}{u}+ \frac{1}{v}=\frac{1}{f}}

We would also have after a detailed considerations of the geometry the following relation valid for a thick lens. Alternatively we will derive the following result from our elegant and powerful matrix method, described here, that we learned in last lecture, lecture — V, here:

\boxed{\frac{1}{f}= (n_l -1) \Big [ \frac{1}{R_1}-\frac{1}{R_2}+\frac{(n_l -1)d_l}{n_lR_1R_2}\Big]}

III. Setting matrix equation for thick lens

1. System matrix for thick lens

Its a cupcake to realize how to draw the paraxial ray diagram for a thick lens system, now that we understood in good detail how we can do it for a general traversal of the light ray in terms of translation and refraction, described here.

So lets just do that. It will be easier to talk as analytical a topic as geometrical optics, when we have a geometrical diagram for it.


Let the ray incident on the lens, strike the first surface at P with coordinates (\lambda_1,\,x_1) and emerge at Q with coordinates (\lambda_2,\,x_2) where \lambda_1   and \lambda_2   are optical direction cosines as we defined in lecture — V, here.

As we can easily see the ray undergoes two refractions — one at P and one at Q, and one translation through thickness t, from P to Q — with a paraxial assumption in medium with refractive index n   .


\begin{pmatrix}\lambda_2 \\ x_2 \end{pmatrix}= \begin{pmatrix}1 & -P_2 \\ 0 & 1 \end{pmatrix} \begin{pmatrix}1 & 0 \\ \frac{t}{n} & 1 \end{pmatrix}\begin{pmatrix}1 & -P_1 \\ 0 & 1 \end{pmatrix} \begin{pmatrix}\lambda_1 \\ x_1 \end{pmatrix}


P_1 = \frac{n-1}{R_1}, \hspace{5pt}P_2 = -\Big (\frac{n-1}{R_2} \Big )= \frac{1-n}{R_2}

we have the system matrix for a thick lens.

\boxed{S=\begin{pmatrix} b & -a\\ -d & c \end{pmatrix} = \begin{pmatrix} 1- \frac{P_2t}{n}&-P_1-P_2 \big[ 1-\frac {tP_1}{n} \big ] \\ \frac{t}{n} & 1- \frac{t}{n}P_1\end{pmatrix}}

2. System matrix for thin lens

A limiting case of thick lens

For a thin lens we can safely assume the limiting thick lens whose thickness goes to zero: t \to 0. This way — by approximating it from the thick lens, we do not lose any generality, that we would lose if a thin lens were to be considered a priori with the condition: t \to 0.

We obtain:

\boxed{S=\begin{pmatrix} 1 & -P_1-P_2 \\ 0 & 1\end{pmatrix}}

So, we have: a = P_1 + P_2,\,\,\, b = c = 1, \,\,\,d = 0.

Now if we have x_1 = 0 then x_2    has to be zero as well. Thats because an image has to be axial, for an axial object point.

So we have \boxed{bD_2 + aD_1D_2 - cD_1 - d = 0, \,\,\,or\,\,\, D_2 + (P_1 + P_2)D_1D_2 - D_1 = 0}.

This implies:

\boxed{\frac{1}{D_2} - \frac{1}{D_1} = P_1 + P_2 = (n-1)\Big (\frac {1}{R_1}-\frac {1}{R_2} \Big )}


\frac{1}{D_2} - \frac{1}{D_1}= \frac{1}{f}, with \hspace {3pt} \hspace {3pt} f = \frac{1}{P_1 + P_2 }= \Big [ (n-1)\Big (\frac {1}{R_1}-\frac {1}{R_2} \Big ) \Big ]^{-1}

Thus we see that the element a in the system matrix is nothing but the inverse of the focal length or power of a lens. We get for thin lens the matrix:

\boxed{\begin{pmatrix} 1 & -\frac {1}{f} \\ 0 & 1\end{pmatrix}}

Also we realize that the power of matrix method is such that we obtain the thin lens equation without even realizing, we did.

\boxed{\frac{1}{f} = (n-1)\Big (\frac {1}{R_1}-\frac {1}{R_2} \Big )}

Its so effortless.

3. Unit and nodal planes of thick lens

From system matrix elements

The elements of the system matrix for a thick lens were obtained in the penultimate section. Lets open the system matrix S that we obtained for a thick lens, and write down these elements one by one.

\boxed{a = P_1 + P_2 \Big ( 1- \frac {t}{n}P_1 \Big ) \,\,\,\,\,\, b= 1 - \frac {t}{n} P_2 \,\, \,\,\,\,\,\, c= 1 - \frac {t}{n} P_1\,\,\,\,\,\, \,\,d= -\frac {t}{n}}

A. Unit planes

The locus of conjugate points — object and image that is, for which linear magnification is unity=1 are called as unit planes. These are also the principal planes that we defined at the beginning of this article, among cardinal points.

A paraxial ray emanating from unit plane in object space will emerge at the same height in image space in the other unit plane. If d_{u1} = 0 and d_{u2} = 0 represent distances of unit planes from refracting surfaces then:

\boxed{b+ad_{u1}=\frac {1}{c-ad_{u2}}= 1}

We note that we would obtain the above result after an equally enlightening analysis of a thick lens system, as we did here and other illuminating analysis in lecture — V, here. This I intend as a homework for the serious student.

Here is the gist — I might as well redo the whole thing, here, and link, but at the moment I will just cite the conditions, that leads to this beautiful result. May be I will give more than that — I just can’t seem to be able to stop myself, but if you work everything out, you will understand more effectively.

Here is how to obtain the above result. Multiply the system matrix that we obtained for thick lens, namely S, with a translation matrix  T_1 on left and a translation matrix T-2   on right, in terms of distances - D_1 and D_2 respectively, as we have seen, these are distances measured from the usual vertices V_1 and V_2 — and the refractive index of air is n_{air}=1 .

We would obtain:

\boxed{\begin{pmatrix} \lambda_2\\x_2 \end{pmatrix} = \begin{pmatrix} b+aD_1 &-a\\ bD_2 + aD_1D_2-cD_1-d & c-aD_2\end{pmatrix} =\begin{pmatrix} \lambda_1\\x_1 \end{pmatrix}}

Note that all that we did here is from our original system matrix S we obtained a new system matrix. The old system matrix S gave the traversal of ray from its coordinate of incidence at first surface, to the coordinate of ray incidence at the second surface.

But before incidence at point P — that is first surface, the ray had traversed from the object point — this is given by matrix T_1 through a translation of distance D_1 and after incidence at point Q — that is at the 2nd surface, the ray would traverse to image point — this is given by matrix T_1 through a translation of distance D_1  .

So including these two additional coordinates changed the old system matrix S to this new one above.

Now let us apply two conditions. One of the conditions we have already used in our lecture today.

A. Condition — i: Axial object points produce axial image points.

That is: x_1 = 0 means x_2=0   . We would get a relationship between D_1   and D_2, namely:

\boxed{bD_2 + aD_1D_2-cD_1-d =0}

This was one of the elements of the matrix we just obtained above, this is the beauty of the matrix method, we obtain important results in terms of the elements of the system matrices. This implies, for the image plane, we obtain the equation:

\boxed{\begin{pmatrix} \lambda_2 \\ x_2 \end{pmatrix} = \begin{pmatrix} b+aD_1 & -a \\ 0 & c-aD_2 \end{pmatrix} = \begin{pmatrix} \lambda_1 \\ x_1 \end{pmatrix}}

Now we apply the second condition.

B. Condition — ii: If our objects points are not axial we get: x_2 = (c - aD_2) x_1 . So magnification of the system is M = \frac{x_2}{x_1} = c - aD_2  . The system matrix obtained above should be unitary, that is, its determinant should be unity=1 .

This gives us what we set out to prove, with the recognition that M=1 in our case, because we are fixed to the unit planes, the ones that have linear magnification by definition, namely:

\boxed{b+ad_{u1}=\frac {1}{c-ad_{u2}}= \frac{1}{M} = 1}

and we have a general result:

\boxed{\begin{pmatrix} \lambda_2 \\ x_2 \end{pmatrix} = \begin{pmatrix} \frac{1}{M} & -a \\ 0 & M \end{pmatrix} = \begin{pmatrix} \lambda_1 \\ x_1 \end{pmatrix}}

Now we have the distances of the unit planes — the primary plane to vertex V_1  distance: d_{u1} and the secondary plane to vertex V_2 distance: d_{u2} can be solved in terms of system matrix elements.

We have:

\boxed{d_{u1}=\frac{1-b}{a}, \hspace{5pt}d_{u2}=\frac{c-1}{a}}

If S is the distance of the object plane from first unit plane and v is the distance of the image plane from second unit plane then;

\boxed{D_1 = u+d_{u1}=u+\frac{1-b}{a}, \hspace{10pt}D_2 = v+d_{u2}=v+\frac{c-1}{a}}

, applying condition — i, that is:

\boxed{bD_2 + aD_1D_2-cD_1-d =0}

and taking 3^{rd} and 4^{th}  term to RHS,

\boxed{D_2 = \frac{d+cD_1}{b+aD_1} \hspace{10pt}det \hspace{2pt}S =bc-ad=1}

so we have:

\boxed{\frac{1}{v}- \frac{1}{u}=a}.

— take this as a homework to prove the above from above steps.

In section — II for thick lens we saw that in stead of a we had the inverse of focal length — the left hand side is equal if we consider the minus sign for sign convention.

So the matrix element again gives the inverse of focal length, as in case of thin lens, which we had seen before. Thus \frac{1}{a} is the focal length of an optical system if distances are measured from unit planes.

Thus for thick lens we have the locations of the unit planes:

\boxed{d_{u1}=\frac{t}{n}{P_2} {\Big [P_1+P_2\Big ( 1-\frac{t}{n}P_1 \Big) \Big]}^{-1} \,\,\,\,\,\,\,\,\,\,\,\, d_{u2}=-\frac{t}{n}{P_1}{\Big [P_1+P_2\Big ( 1-\frac{t}{n}P_1 \Big) \Big]}^{-1}}

For symmetric thick-lens:

\boxed{d_{u1}=\frac{t}{2n} \hspace{10pt}d_{u2}=-\frac{t}{2n}\,\, \,\, \,\,\,\, \,\, \,\, \frac{1}{f}=a = P_1 + P_2 \Big ( 1- \frac{t}{n}P_1\Big) =(n-1)\Bigg [ \frac{1}{R_1}-\frac{1}{R_2}+\frac{(n-1)t}{nR_1R_2}\Bigg]}

This is exactly what we had stated in the beginning of our article in section — II. We see that we can obtain the fundamental equations of thick lens and thin lens almost effortlessly if we are to follow the magic of matrix method.

B. Nodal planes

We apply similar reasoning to obtain the nodal plane distances. By using our condition — ii

\boxed{\begin{pmatrix} \lambda_2 \\ x_2 \end{pmatrix} = \begin{pmatrix} b+aD_1 & -a \\ 0 & c-aD_2 \end{pmatrix} = \begin{pmatrix} \lambda_1 \\ x_1 \end{pmatrix}}

we see that;

\boxed{d_{n1}=\frac{1-b}{a} = d_{u1}}

this is because the media on both sides of the lens is air and also lens is symmetric. We had stated this earlier, now we have evidence.

Similarly second nodal plane is same as second unit plane.


\boxed{d_{n2}=\frac{c-1}{a} = d_{u2}}

4. System of two thin lenses

From system matrix elements of thick lens

If two thin lenses of focal lengths f_1 and f_2 are considered then the system matrix is easy to guess based on our discussion above; a is inverse of focal length. The distance of separation between them is simply a translation of thickness t.

We have,

\boxed{S = \begin{pmatrix}1 & -\frac{1}{f_2} \\ 0 & 1 \end{pmatrix} \begin{pmatrix}1 & 0 \\ t & 1 \end{pmatrix}\begin{pmatrix}1 & -\frac{1}{f_1} \\ 0 & 1 \end{pmatrix}}

which yields the matrix elements:

\boxed{a=\frac{1}{f_1}+\frac{1}{f_2}-\frac{t}{f_1f_2}, \,\,\,\,\,\,\,\, b=1-\frac{t}{f_2}, \,\,\,\,\,\,\,\, c=1-\frac{t}{f_1}, \,\,\,\,\,\,\,\, d = -t}

and we have


This is the result we had used in studying achromatic doublets, here. Also the unit planes are located at

\boxed{d_{u1}=\frac{1-b}{a}=\frac{tf}{f_2}\, \, \, \, \, \, \, \,  \, \, \, \, d_{u2}=\frac{c-1}{a}=-\frac{tf}{f_1}}

5. System of two thin lenses

Ramsden and Huygens’s eyepieces

This is merely an application of section – 4 above. I leave this to your good sense for trying this as a homework problem and get some conceptual clarity on the differences between the two types of eye pieces.

Categories: Courses I developed, matrix, optics, Physics, Teaching

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