Thick lenses systems

The optical matrix formalism is applied to find parameters such as focal distance, back and front focal points, principal planes, and the equation relating object and image distances for a thick spherical lens immerse in air. Then, the formalism is applied to systems compound of two, three and N thick lenses in cascade. It is found that a simple Gaussian equation is enough to relate object and image distances no matter the number of lenses.


INTRODUCTION
The matrix optics formalism is a powerful tool that can be applied to the study of thick multi-lenses systems. It allows an intuitive approach in order to understand more deeply dispositive such as cameras, that may consist of several thick lenses in cascade. On the other hand, when the object-image equation for a single thick lens is deduced using matrix formalism, an expression that is complicated to deal with is obtained. However, through a coordinate transformation on the image and object distances, it is possible to obtain a very interesting result which simplifies the expression to a formula that is formally identical to the familiar expression used for thin lenses, that is, the Gaussian equation. In this work, such expression introducing optical parameters such as effective focal distance, front and back frontal length, and principal planes was fully derived. This is not a new result, as it has been pointed out in Hecht (2017), Jenkins and White (2001), and Born and Wolf (1999). Nevertheless, in this work this feature is demonstrated in a different way, by using the matrix formalism (not used in the references before mentioned for this specific result), which allows exploration of other situations. Further, this analysis was extended to a system formed by two thick lenses separated by a distance d which gave an interesting result that all optical parameters characterizing this system and the object-image equation are mathematically identical to the previously found expressions for a single thick lens. Furthermore, this analysis was extended to optical systems comprising three and N thick lenses, and formal identical results were once again found. Indeed, one of the more interesting results in this work is this generalization and, as a direct consequence, the possibility of expressing the relation between object and image distances for an arbitrary number of thick lenses in cascade just in a single equation. Actually, this general result is predicted by Feynman (2006) without demonstration.

MATRIX OPTICS
The optical matrices used for this study, that is, the refraction and displacement matrices, can be found in Hecht (2017). An important remark is that these matrixes are deduced for the paraxial approximation, making all our results to be valid in that case. In this paper, these results will be briefly derived and the interpretation and further application of these matrices to develop the expressions for spherical thick lenses will be shown. Also, the classical signal convention, which states that object distances to the left (right) of some reference interface will be positive (negative), whereas image distances placed to the right (left) of such interface will be positive (negative) was adopted. As for the curvature radios of surfaces, they will be positive (negative), if they were convex (concave). This convention is currently adopted in practically all specialized literature.
Let us suppose that an optical ray that makes an angle α 1 with respect to horizontal direction z (from here on, the optical axis of the system), is propagated in an optical medium with a refraction index n 1 . It then hits a spherical medium with refraction index n 2 at a height y on the surface, measured from the optical axis, suffers refraction and, as a consequence, changes the angle with respect to the optical axes to α 2 ( Figure 1).
In Figure 1, 'R' denotes the radium of the sphere. The equation that completely describes refraction is: (1) Note that in Equation 1, if 'R' tends to infinity, then the equation reduces to Snell's law (in paraxial form), as expected. We can express Equation 1 in a matrix form, by defining the vector (nα, y):  corresponding to the n 2 α 2 component reproduces Equation 1. The other term, y 2 = y 1 , simply describes that no change in height was verified to the point of refraction. Now we describe the change of height, that is, in 'y' coordinate of the optical ray, as it propagates in an optical medium ( Figure 2).
It is easy to see that, as the ray propagates a distance d measured over the optical axis in the medium with refraction index n 1 , the changes in y coordinate is given by: ( As we work with the paraxial approximation, that is, α 1 << 1, so, tanα 1 ~ α 1 , Equation 3 becomes: (4) Now, we can write Equation 4 in matrix form as: Performing the product, the expression that corresponds to the element 'y 2 ', reproduces Equation 4, while the result corresponding to the component n 2 α 2 reproduces the Snell law, in paraxial form, at the interface between the two medium.

THICK LENSES IN AIR, FOCAL POINTS AND PRINCIPAL PLANES
The vantage of the matrix formalism is that it can be applied to several optical mediums in cascade by simply multiplying the matrixes corresponding to each element. Now, it will be applied to a spherical thick lens immersed in air, which consist in two refractive spherical surfaces of radios R 1 and R 2 , separated by a distance d over the optical axis, which is the thickness of the lens, as shown in Figure 3. In Figure 3, V 1 and V 2 are vertices points, localized at the intersections of the spherical surfaces of radio R 1 and R 2 , with the optical axis, respectively. Also, it can be seen that an optical ray making an α 1 angle with respect to the optical axis is refracted by the first surface, at a height y 1 , which propagates inside the lens (a distance d measured over the optical axis) and is then refracted by the second surface, leaving the lens with an angle of α 2 at a height y 2 with respect to the optical axis. In order to mathematically describe the journey of the optical ray through the lens, it can be written as: As the lens is immersed in air, the indexes of refraction that multiply the angles in the vector components are both equal to one. Also, the elements corresponding to file 1 column 2 in both of the refraction matrixes in Equation 6 can be easily understood by looking to the general form of this element in Equation 2, where its numerator could be described in words as: "(minus) index of refraction to the right of the interface minus the index of refraction to the left of the interface (over R)". Then, the first interface (with radius 'R 1 ') is surrounded by air (index n ar =1) to the left and glass (of index n) to the right. The same applies to the second surface (with radius 'R 2 ').
A convenient way to look at Equation 6 is from right to left: first, we have the input ray, hitting the lens with angle α 1 at a height y 1 , it is refracted by the first surface, with radio R 1 , then the ray is displaced a horizontal distance d, inside the lens, which has a refraction index n, and finally it is refracted by the second refractive surface, with radius R 2 , and this gave the resultant optical ray leaving the lens, characterized by α 2 and y 2 . Recalling that matrix product is not commutative, the correct order of the matrixes is fundamental in order to achieve the right

result.
Performing the usual operation of matrix product, Equation 6 results: Note that, if d = 0, that is, when the thick lens becomes a thin lens, the element a 12 in the matrix characterizing the lens, becomes equal to minus the inverse of the focal distance for a thin lens. From here on, this term will be written as -1/f (parameter f for thick lens will be analyzed later on). Now, the back focal length, z b , that is, the distance, measured from V 2 , to which a thick lens focalizes a bunch of optical ray incident parallel to the optical axis, will be deduced ( Figure 4).
Equivalently, it may be though as a plane wave front (being the wave fronts perpendicular to the optical rays) that is incident in the length, and, after refraction, the lens produced a convergent spherical wave, whose center is at back focal point. Some of these fronts are shown in Figure 4. Now, the expression to find the height y, over the optical axis, of an optical ray that has travelled a horizontal distance z after leaving the lens will be written. In order to mathematically describe the whole process, the displacement matrix in air must be used in addition to the thick lens matrix in the following way: left, that is, an input ray (α 1 y 1 ) is refracted by the lens, and then propagates an horizontal distance z. Performing the product of the matrixes, Equation 8 becomes: Now, the expression for y: Noting that in our particular case α 1 =0, we seek for that value of z that makes y null for all values of y 1 (Figure 4). This is by definition, the back focal point, z b : Note that, if d=0, z b =f, that is, the same result for thin lenses, as expected (and in this particular case, of course, the focal distance became the familiar expression for thin lenses).
A consideration of signal must be mentioned here. Back focal length is obviously an image point, and the corresponding signal convention should be applied, that is, if the result is positive, it should be localized to the right of V 2 , as schematically shown in Figure 4, where a positive value of z b is assumed. If negative, it should be localized to the left of V 2 . Now, we can deduce the front focal length, z f , that is the distance measured from V 1 on the optical axis, in which an object point (or a source of spherical divergent wave fronts), must be located in order that a thick lens produce a bunch of parallel rays after refraction (corresponding to plane wave fronts). The situation is shown in Figure 5. Now we write the expression to find (α 2, y 2 ): indicating that a ray with height y and angle α 1 is propagated at a certain distance z, before it hits the length. Performing the product indicated in Equation 12: Now, we analyze the term corresponding to α 2 from We must impose the conditions y = 0, and α 2 = 0, which are compatible with an object point over the optical axis producing, via the thick lens, an "image at infinity" (or a bunch of parallel rays), as indicated in Figure 5. As a consequence, the term between parentheses in Equation 14 should be equal to zero, for any value of α 1 . The particular value of z that makes this condition be fulfilled is the z f : If we set d = 0, that is, a thin lens, Equation 15 reduces, as expected, to the expression for thin lenses. Front focal point is an object point and, if positive, should be located to the left of V 1 (as indicated in Figure  5, where that case was supposed). If it has negative signal, of course, it should be localized to the right of V 1 .
We can now investigate the significance of f for thick lenses. In Figure 6, we see the front focal point for a given thick lens, and a generic optical ray, with α 1 angle, entering the length at y 1 and leaving the lens at y 2 , with α 2 = 0 (definition of z f ). We now project back the output ray until it intersects the projection of input ray. Doing this for every possible α 1 angle of those optical rays departing from z f being refracted by the thick lens, we obtain a plane, perpendicular to the optical axis, at a distance x from z f . This plane is called principal plane. In order to find the distance x, we observe that we can express the angle α 1 , in the paraxial approximation as: Then we express x as: To find y 2 we write from Equation 7: Again, we must write the expression for α 1 in the paraxial approximation: And obtain: (20) Now, using Equation 20 , and with a little algebra, it is easy to find that: In this way, we observed that the distance from z f to the principal plane is given by f. We can also generate another plane, using the concept of back focal point in a totally symmetric way, and it can be found that the distance between this second plane and z b is also f (Figure 7). These two planes are called first principal plane (1PP), and second principal plane (2PP), and they can be seen in Figure 8. It should be noted that the principal planes are seldom inside the lens, as shown in Figure 8. Depending on the specific lens data, they could be outside the lens.
For reasons that will be subsequently presented, it is convenient to localize the 1PP and 2PP with respect to the V 1 and V 2 vertices points, respectively; we call these distances h 1 and h 2 (Figure 8): From the Equation 24, it can be seen that distance h 1 obeys the following signal convention: if positive, it is located to the right of V 1 , but if negative, to the left of V 1 . For h 2 (Equation 25), the signal convention for images with respect to V 2 holds, that is, if positive, h 2 must be located to the right of V 2 and if negative, to the left of V 2.

THICK LENS IN AIR AND IMAGE CONDITION
Now, the image condition for a thick lens in air will be deduced. We consider an object, with height h o to a distance s o from V 1 . An image will be formed by the thick lens at distance s i , measured with respect to V 2 , and  height h i . The situation is visualized in Figure 9.
We see then that optical rays from object points will propagate a distance s o over the optical axis; it will then be refracted by the thick lens and propagate a distance s i , to the corresponding image point. In mathematical terms, we can write: In order to mathematically find the image condition, we can consider an object point, for example, the one at the top of the object. This point is a front of spherical divergent waves (some of which are shown in Figure 9, perpendicular to the optical rays). The lens, in order to form an image, should convert this divergent wave front into a convergent one, whose convergence point is the image point corresponding to the object point at the top of the object. Then, we write, from Equation 31, the expression corresponding to h i : It is easy to see that the height of a specific image point (for example the one corresponding to the top), should not depend on the angle α 0 . In this way, the image condition is: From Equation 38, it is easy to see that, in order to obtain an expression that looks like the Gaussian formula for thin lens, the following conditions should be imposed: ( Equations 42 and 43 make Equation 41 to be fulfilled (this demonstration requires a certain amount of algebra, and it is shown in the Appendix). As a consequence, we were able to find new object and image distances that allow us to write Equation 35  (45) These expressions are closely related to Equations 24 and 25, that is, the distances of the principal planes to their respective vertices points. As the distances s o and s i are measured with respect to the vertices, this means that, in order to obtain a simpler lens equation, we must measure the object and image distances with respect to the 1PP and 2PP respectively, as shown in Figure 10. As a consequence, we can replace the thick lens by the principal planes, and write the Gauss equation: Of course, if we make d=0, that is, a thin lens, the two principal planes coalesce into one, and in that case, s o = s´o and s i = s´i.

TWO THICK LENSES SYSTEMS
The formalism developed in the last sections can be generalized to a two lenses system.
Let us suppose a system of two thick lenses, characterized by matrixes A and B, and separated by a distance d. The lens A has refraction index n a , thickness d a , radius R 1a , and R 2a whereas for the lens B, index n b , thickness d b , radius R 1b , and R 2b . Matrix elements for each lens are the same as in Equations 27 to 30, with the specific parameters for each lens: (47) (50) The system is shown in Figure 11. Note that we have now four vertices points, V 1a , V 2a , V 1b , and V 2b .
We write for the matrix of the two lenses system: (59) It may be illustrative to develop the element M ab12 of the compound system using the matrix elements explicitly (Equations 47 to 54). As already seen for one single lens, the element "1-2" is related to the focal distance. After a little algebra, Equation 57 becomes: It will be shown that this expression corresponds to the negative of the inverse of the focal distance for the two thick lenses, separated by a distance d. Here, it is referred to as focal distance, f ab . Meanwhile, it is easy to see that, making d a = d b = 0, that is, two thin lenses, Equation 60 reduces to the well-known expression of the (negative of the inverse) focal distance for two thin lenses separated by a distance d (where f a and f b , of course, are reduced to the expressions of focal distances for thin lenses).
We can determine the back focal length for this system, proceeding in an analogous way as we did for the single thick lens (Equation 8), that is, As we already know, for back focal point, α 1 =0, and we must compute the value of z that makes y null for all values of y 1 : Writing M ab12 as -1/f ab , so: Now, we take a look at the expression for z b of the single thick lens (Equation 11). It may be written, in term of the matrix element of the lens, as z b = fa 22 .
Also, we can find the front focal length for the compound system of thick lenses, using a reasoning analogous to that used in Equations 12 to 15, the result is: Once more, we observe that the expression is formally identical to the corresponding front focal length for a single thick lens (Equation 15), that may be written using the matrix element as z f = fa 11 . The next logical step is to find the principal planes for the two thick lens system and find the relation with the focal distance, f ab (Figure 12).
In Figure 12, we find the 1PP for the two thick lens system using, as before, the property of front focal point, zf, and projecting back the y 2 coordinate. Thereafter, we write for the angle α 1 (in the paraxial approximation): Then, we can write for x: (67) which is, of course, formally the same equation obtained when the single thick lens was studied. Now, in order to obtain an expression for y 2 in our two thick lenses systems, we write: Note that this expression allows us to obtain the height and angle, α 2 and y 2 , of a ray leaving the compound system at the last surface (R 2b ), as a function of the corresponding parameters α 1 and y 1 of a ray hitting the systems at the first surface (R 1a ).

From Equation 68
, we obtain for y 2 : Our next task is to demonstrate that x = f ab . It is a long algebra but also, one of the principal points of this work, so it will be developed here. We write Equation 72 explicitly as a function of the matrix elements: We see, in Equation 74, that the term within the parentheses, multiplying the a 11 db 22 factor, is the M ab12 matrix element (Equation 57), which is equal to -1/ f ab . We write then: ) (   22  22  21  12  12  22  21  12  12  21  21  12  12  21  11  12  22  22  11  11  22  11  12  21  11  11  21  22  11  22  11 We can express the factors between parenthesis in the second and sixth terms using the fact that: -1/f ab =b 11 a 12 +b 12 (da 12 + a 22 ), then: Now, the elements a ij and b ij must be explicitly replaced in Equation 79, using Equations 47 to 54. Working with the first factor: .
After a simple algebra, it is obtained: can be finally written as: With this demonstration, it has been shown that the interpretation of f ab as the focal distance of the two thick lenses systems is correct, and totally analogous to the case of one single thick lens. We can now write the expression for focal distance of two thick lenses separated by a distance d as: The principal plane found and shown in Figure 12 is the first principal plane. Of course, the second principal plane is easily found and is verified that its distance to the back focal point, z b , is also f ab .
The next steps now are, in close analogy with the single thick lens case, to find the distances of the 1PP and 2PP, h 1 and h 2 , which will be measured with respect to the outermost vertices points of the system, that is, V 1a and V 2b respectively, and also determine the image condition.
In order to find the distances of the principal planes with respect to the outermost vertices points, we write: (85) ( ) Where Equations 64 and 65 were used to express front and back focal distances for the two thick lens system. As before, h 1 and h 2 follow the signal convention for object and images, respectively. To find image condition, we measure the object distance, s o , with respect to V 1a and image distance, s i , with respect to V 2b (Figure 13).
Writing the matrix equation for the object-two thick lenses system-image: which is formally identical to Equation 35; therefore, we can now follow the same steps of Equations 36 to 45, to the extent of writing the image condition for the two thick lenses system in the simple form of a Gaussian equation. To achieve this, of course, object and image distances must be measured with respect to the principal planes. With this condition, a complicated system as shown in Figure 13 is reduced to the one shown in Figure 14. Furthermore, that system is described analytically by: (89)

THREE AND N THICK LENSES, IMAGE CONDITION
The expressions for three, four and N thick lenses systems follow the same general structure. Consider a three thick lenses system, A, B, and C, being the a, b lenses identical to the system seen in the last section, and C lens is located to a distance d bc to the right of B lens. The c lens, of course has matrix elements mathematically identical to those represented in the Equations 47 to 50, and characterized with usual parameters, refraction index n c , thickness d c , radius R 1c , and R 2c. The matrix for this system is: front focal point, back focal point, etcs, can be obtained from these elements, using the expressions already seen for the two thick lenses system, and substituting the corresponding matrix element to the three lenses system. Of course, the focal distance is associated to the M abc12 element as usual. A closer look at Equation 91 allows us to write them in a compact form: ( ) analysis to three and N thick lenses systems, analogous results was found. It is interesting to observe that all well-known expressions relating thin lenses systems can be deduced from the more realistic formulas derived in this work, simply by making the parameter d a of the lens equal to zero, which is, passing from a thick to a thin lens.
This result is of particular interest given that in an optical system with an arbitrary number of lenses, allow us to find a single pair of principal planes associated to that system, and determine image distances (given the object distances) by resolving a simple Gaussian equation. This last result that was mentioned by Feynman (2006) in his lectures book, although well known to all specialists in the field, had not been formally derived in the basic or intermediate optic literature to the best of the researcher's knowledge.

APPENDIX
Our task here is to demonstrate that the following identity is true: Clearly, the term between brackets in Equation (A5) is -1/f a , and the identity is verified.