# Optics Series Lecture, Lecture – VI.

“Application of Matrix Method to Thick Lenses.” This lecture was delivered on 7th February 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.

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

We assumed a paraxial system.
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 lecture-I and lecture-II.
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 you define 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 — lecture – 5 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.

A thick lens can be considered to be an equivalent of multiple thin lenses. Let us draw two suitable thick lens diagrams.

## I. Cardinal Points.

Lets first discuss the parameters of the thick lens system.

First Focal Point or Object Focus, Fo.
Rays emerging from Fo, incident on the first spherical surface, will emerge parallel to Optical Axis. Check the first diagram above.
Second Focal Point or Image Focus, Fi.
If rays incident on first spherical surface are parallel then they meet at Fafter refraction from second spherical surface. Check the 2nd diagram above.
F.F.L.
The Forward Focal Length is the distance between object focus Fand vertex to the first spherical surface, V.
B.F.L.
The Backward Focal Length is the distance between image focus Fand vertex to second spherical surface, V.
Image Surfaces:
when rays from Fand F— object and image focii, are extrapolated they meet on a curved surface rather than a 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 Fand Fare extrapolated. The Principal Planes meet the optical axis at 1st and 2nd Principal Points H1 and H2.
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 N1 and N2 on optical axis. N1 is called as First Nodal Point and N2 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, Fand F, Principal Points H1 and Hand Nodal Points N1 and N2, are together known as Cardinal Points of the system.
The length between H1 and His nearly equal to 1/3 length between Vand V2 for ordinary glass lenses in air.

## II. Thick Lens Equations.

Thick lenses are equivalent to two spherical refracting surfaces separated by a distance d 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 (H1)  and image distance v is measured from Second Principal Plane (H2): $\inline&space;\fn_cs&space;\frac{1}{u}+&space;\frac{1}{v}=\frac{1}{f}$ 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 that we learned in last lecture, lecture-5$\inline&space;\fn_cs&space;\frac{1}{f}=&space;(n_l&space;-1)&space;\Big&space;[&space;\frac{1}{R_1}-\frac{1}{R_2}+\frac{(n_l&space;-1)d_l}{n_lR_1R_2}\Big]$

## 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. 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 (λ1, x1) and emerge at Q with coordinates (λ2, x2) where λ1 and λ2 are optical direction cosines as we defined in lecture – 5. As we can easily see the ray undergoes two refraction — at P and at Q, and one translation through thickness t, from P to Q — with a paraxial assumption in medium with refractive index n. Accordingly: $\inline&space;\fn_cs&space;\begin{pmatrix}\lambda_2&space;\\&space;x_2&space;\end{pmatrix}=&space;\begin{pmatrix}1&space;&&space;-P_2&space;\\&space;0&space;&&space;1&space;\end{pmatrix}&space;\begin{pmatrix}1&space;&&space;0&space;\\&space;\frac{t}{n}&space;&&space;1&space;\end{pmatrix}\begin{pmatrix}1&space;&&space;-P_1&space;\\&space;0&space;&&space;1&space;\end{pmatrix}&space;\begin{pmatrix}\lambda_1&space;\\&space;x_1&space;\end{pmatrix}$ With $\inline&space;\fn_cs&space;P_1&space;=&space;\frac{n-1}{R_1},&space;\hspace{5pt}P_2&space;=&space;-\Big&space;(\frac{n-1}{R_2}&space;\Big&space;)=&space;\frac{1-n}{R_2}$ we have the system matrix for a thick lens. $\inline&space;\fn_cs&space;S=\begin{pmatrix}&space;b&space;&&space;-a\\&space;-d&space;&&space;c&space;\end{pmatrix}&space;=&space;\begin{pmatrix}&space;1-&space;\frac{P_2t}{n}&-P_1-P_2&space;\big[&space;1-\frac&space;{tP_1}{n}&space;\big&space;]&space;\\&space;\frac{t}{n}&space;&&space;1-&space;\frac{t}{n}P_1\end{pmatrix}$

### 2. System Matrix for Thin Lens As A Limiting Case Of Thick Lens.

For a thin lens we can safely assume the limiting thick lens whose thickness goes to zero: t → 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 → 0. We obtain: $\inline&space;\fn_cs&space;S=\begin{pmatrix}&space;1&space;&&space;-P_1-P_2&space;\\&space;0&space;&&space;1\end{pmatrix}$. So, we have: a = P1 + P2, b = c = 1, d = 0. Now if we have x1 = 0 then x2 has to be zero as well. Thats because an image has to be axial, for an axial object point. So we have bD2 + aD1D2 – cD1 – d = 0or; D2 + (P1 + P2)D1D2 – D1 = 0. This implies: $\inline&space;\fn_cs&space;\frac{1}{D_2}&space;-&space;\frac{1}{D_1}&space;=&space;P_1&space;+&space;P_2&space;=&space;(n-1)\Big&space;(\frac&space;{1}{R_1}-\frac&space;{1}{R_2}&space;\Big&space;)$or $\inline&space;\fn_cs&space;\frac{1}{D_2}&space;-&space;\frac{1}{D_1}=&space;\frac{1}{f},&space;with&space;\hspace&space;{3pt}&space;\hspace&space;{3pt}&space;f&space;=&space;\frac{1}{P_1&space;+&space;P_2&space;}=&space;\Big&space;[&space;(n-1)\Big&space;(\frac&space;{1}{R_1}-\frac&space;{1}{R_2}&space;\Big&space;)&space;\Big&space;]^{-1}$ Thus we see that the element a in the system matrix is nothing but inverse of the focal length or Power of a lens. We get for thin lens the matrix: $\inline&space;\fn_cs&space;\begin{pmatrix}&space;1&space;&&space;-\frac&space;{1}{f}&space;\\&space;0&space;&&space;1\end{pmatrix}$ Also we realize that the power of matrix method is such that we obtain the thin lens equation without even realizing. $\inline&space;\fn_cs&space;\frac{1}{f}&space;=&space;(n-1)\Big&space;(\frac&space;{1}{R_1}-\frac&space;{1}{R_2}&space;\Big&space;)$ 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, to write down these elements one by one. $\inline&space;\fn_cs&space;a&space;=&space;P_1&space;+&space;P_2&space;\Big&space;(&space;1-&space;\frac&space;{t}{n}P_1&space;\Big&space;)&space;\\&space;b=&space;1&space;-&space;\frac&space;{t}{n}&space;P_2&space;,&space;\hspace&space;{3pt}&space;c=&space;1&space;-&space;\frac&space;{t}{n}&space;P_1,&space;\\d=&space;-\frac&space;{t}{n}$

#### A. Unit Planes.

The locus of conjugate points — object and image that is, for which linear magnification is 1 are called as Unit Planes. These are also the Principal Planes that we defined at the beginning of this article. A paraxial ray emanating from unit plane in object space will emerge at the same height in image space in the other unit plane plane. If du1 and du2 represent distances of unit planes from refracting surfaces then: $\inline&space;\fn_cs&space;b+ad_{u1}=\frac&space;{1}{c-ad_{u2}}=&space;1$

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 – 5. 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 T1 on left and a translation matrix T2 on right, in terms of distances -D1 and D2 respectively, as we have seen, these are distances measured from the usual vertices V1 and V2 — and the refractive index of air is 1. You would obtain: $\inline&space;\fn_cs&space;\begin{pmatrix}&space;\lambda_2\\x_2&space;\end{pmatrix}&space;=&space;\begin{pmatrix}&space;b+aD_1&space;&-a\\&space;bD_2&space;+&space;aD_1D_2-cD_1-d&space;&&space;c-aD_2\end{pmatrix}&space;=\begin{pmatrix}&space;\lambda_1\\x_1&space;\end{pmatrix}$Note that all we did here is from our original system matrix 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 coordinate of ray incidence  at 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 T1 through a translation of distance D1) and after incidence at point Q (that is at 2nd surface) the ray would traverse to image point (this is given by matrix T1 through a translation of distance D1). So including these two additional coordinate 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. 1. Condition – i: Axial object points produce axial image points. That is: x1 = 0 means x2 = 0. We would get a relationship between D1 and D2, namely: $\inline&space;\fn_cs&space;bD_2&space;+&space;aD_1D_2-cD_1-d&space;=0$. 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: $\inline&space;\fn_cs&space;\begin{pmatrix}&space;\lambda_2&space;\\&space;x_2&space;\end{pmatrix}&space;=&space;\begin{pmatrix}&space;b+aD_1&space;&&space;-a&space;\\&space;0&space;&&space;c-aD_2&space;\end{pmatrix}&space;=&space;\begin{pmatrix}&space;\lambda_1&space;\\&space;x_1&space;\end{pmatrix}$ Now we apply the second condition. 2. Condition – ii: If our objects points are not axial we get: x2 = (c – aD2)x1. So magnification of the system is M = x2 / x1 = c – aD2. The system matrix obtained above should be unitary, that is, its determinant should be 1. This gives us what we set out to prove, with the recognition that M = 1 in our case, because we are fixated to the unit planes, the ones that have linear magnification by definition, namely: $\inline&space;\fn_cs&space;b+ad_{u1}=\frac&space;{1}{c-ad_{u2}}=&space;\frac{1}{M}&space;=&space;1$ and we have a general result: $\inline&space;\fn_cs&space;\begin{pmatrix}&space;\lambda_2&space;\\&space;x_2&space;\end{pmatrix}&space;=&space;\begin{pmatrix}&space;\frac{1}{M}&space;&&space;-a&space;\\&space;0&space;&&space;M&space;\end{pmatrix}&space;=&space;\begin{pmatrix}&space;\lambda_1&space;\\&space;x_1&space;\end{pmatrix}$

Now we have the distances of the unit planes — the primary plane to vertex V1 distance: du1 and the secondary plane to vertex V2 distance: du2 can be solved in terms of system matrix elements. We have: $\fn_cs&space;d_{u1}=\frac{1-b}{a},&space;\hspace{5pt}d_{u2}=\frac{c-1}{a}$If u 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;$\fn_cs&space;D_1&space;=&space;u+d_{u1}=u+\frac{1-b}{a},&space;\hspace{10pt}D_2&space;=&space;v+d_{u2}=v+\frac{c-1}{a}$, applying condition – i , that is:$\inline&space;\fn_cs&space;bD_2&space;+&space;aD_1D_2-cD_1-d&space;=0$and taking 3rd and 4th term to RHS,$\fn_cs&space;D_2&space;=&space;\frac{d+cD_1}{b+aD_1}&space;\hspace{10pt}det&space;\hspace{2pt}S&space;=bc-ad=1$so we have — take this as a homework to prove the following from above few steps: $\fn_cs&space;\frac{1}{v}-&space;\frac{1}{u}=a$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 1/a is the focal length of a optical system if distances are measured from unit planes.

Thus for thick lens we have the locations of the unit planes: $\fn_cs&space;d_{u1}=\frac{P_2t}{n}\frac{1}{\Big&space;[P_1+P_2\Big&space;(&space;1-\frac{t}{n}P_1&space;\Big)&space;\Big]}&space;\\&space;d_{u2}=-\frac{t}{n}\frac{P_1}{\Big&space;[P_1+P_2\Big&space;(&space;1-\frac{t}{n}P_1&space;\Big)&space;\Big]}$For symmetric thick-lens: $\fn_cs&space;d_{u1}=\frac{t}{2n}&space;\hspace{10pt}d_{u2}=-\frac{t}{2n}\\&space;\\&space;\\&space;\frac{1}{f}=a&space;=&space;P_1&space;+&space;P_2&space;\Big&space;(&space;1-&space;\frac{t}{n}P_1\Big)\\&space;\\&space;\\&space;=(n-1)\Big&space;[&space;\frac{1}{R_1}-\frac{1}{R_2}+\frac{(n-1)t}{nR_1R_2}\Big]$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 $\inline&space;\fn_cs&space;\begin{pmatrix}&space;\lambda_2&space;\\&space;x_2&space;\end{pmatrix}&space;=&space;\begin{pmatrix}&space;b+aD_1&space;&&space;-a&space;\\&space;0&space;&&space;c-aD_2&space;\end{pmatrix}&space;=&space;\begin{pmatrix}&space;\lambda_1&space;\\&space;x_1&space;\end{pmatrix}$we see that; $\fn_cs&space;d_{n1}=\frac{1-b}{a}&space;=&space;d_{u1}$ 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. So: $\fn_cs&space;d_{n2}=\frac{c-1}{a}&space;=&space;d_{u2}$

### 4. System of two thin Lenses From System Matrix Elements of thick lens.

If two thin lenses of focal lengths f1 and f2 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, $\fn_cs&space;S&space;=&space;\begin{pmatrix}1&space;&&space;-\frac{1}{f_2}&space;\\&space;0&space;&&space;1&space;\end{pmatrix}&space;\begin{pmatrix}1&space;&&space;0&space;\\&space;t&space;&&space;1&space;\end{pmatrix}\begin{pmatrix}1&space;&&space;-\frac{1}{f_1}&space;\\&space;0&space;&&space;1&space;\end{pmatrix}$which yields the matrix elements: $\fn_cs&space;a=\frac{1}{f_1}+\frac{1}{f_2}-\frac{t}{f_1f_2},&space;\hspace{10pt}b=1-\frac{t}{f_2},&space;\\\hspace{10pt}c=1-\frac{t}{f_1},&space;\hspace{10pt}d&space;=&space;-t.$ and we have $\fn_cs&space;a=\frac{1}{f}=\frac{1}{f_1}+\frac{1}{f_2}-\frac{t}{f_1f_2}$This is the result we had used in studying achromatic doublets. Also the unit planes are located at $\inline&space;\fn_cs&space;d_{u1}=\frac{1-b}{a}=\frac{tf}{f_2}\hspace{8pt}d_{u2}=\frac{c-1}{a}=-\frac{tf}{f_1}$.

### 5. System of two thin Lenses, Ramsden and Huygen’s eyepieces.

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

This 6th, 1 and 1/2 hour lecture, actually finishes our unit-I of paper-IV, which is an Optics Honors course. The 1 hour residue (out of purported 10 hours for the unit of 10 lectures of 1 hour duration)  is for reasons of small left out part (as homework) and/or efficiency in teaching this unit.  We will later compare this with second unit of this paper and 1st unit of Electricity and Magnetism,  which will follow as a sequence and these units are uniformly distributed as 10 hour lecture-units.