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enter image description here

Suppose we have the schematic above. The first lens has a focal length of 80mm and the second one has a focal length of 60mm, and both are biconvex. My job is to determine, graphically the equivalent focal length.

As two thin lenses are equivalent to one thin lenses, my attempt was to find the principal planes and then infer, from the isochronicity of light rays to see where rays intercept and get the equivalent focal length. But I have thought about this for a long tima now and am stuck. Anyone can give a hand?

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  • $\begingroup$ When you sa that you want to determine the focal length graphically do you mean (a) by ray tracing alone or (b) by a combination of ray tracing and plotting a graph or (c) making experimental measurements and plotting a graph? $\endgroup$ Commented Mar 16, 2020 at 0:40
  • $\begingroup$ I was thinking about ray tracing alone, but that is merely because I have no idea what a graph means in this case... $\endgroup$
    – Bidon
    Commented Mar 16, 2020 at 8:54
  • $\begingroup$ BTW two thin lenses are only equivalent to another thin lens if they are placed in contact or are separated by a distance d which is small compared with the focal lengths f of both lenses. Their principal planes are then assumed to coincide. But if their separation d is comparable with f then the principal planes do not coincide. $\endgroup$ Commented Mar 16, 2020 at 18:23

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A compound lens is characterised not only by an effective focal length (EFL) but also by the positions of front and back principal planes (FPP, BPP). The latter do not usually coincide with the vertices of the initial and final refracting surfaces. Consequently you require 3 pairs of object and image positions to determine all 3 unknown parameters (EFL, FPP, BPP).

The effect on ray tracing of the separation of the principal plane into two is that the height and angle at which a ray is incident on the 1st principal plane are transferred directly to the 2nd principal plane, as though the intervening space does not exist.

enter image description here

It is straightforward to construct object and image positions for the compound lens using the intermediate image formed by the first lens $L_1$. This image becomes the object for the second lens $L_2$. The method for constructing the final image formed by two thin lenses is illustrated in the diagram below.

enter image description here

Ray 1 from the object is parallel to the axis. Is refracted at lens $L_1$ and passes through the back focal point $F_1'$ of this lens. Ray 2 passes through the pole (= vertex) of this lens and is not refracted. The intersection of rays 1 & 2 gives the position of the intermediate image $I'$ of object $O$ formed by lens $L_1$.

Now trace ray 3 from $O$ through the front focal point $F_1$ of lens $L_1$. This is refracted parallel to the optical axis. When it reaches lens $L_2$ it is refracted through the back focal point $F_2'$ of this lens. Trace ray 4 back from $I'$ through the pole of lens $L_2$. Where rays 3 & 4 intersect is the final image point $I$.

This method can of course be extended for a system of any number of thin lenses or refracting surfaces, constructing the intermediate image points for each lens or surface in the order in which they are encountered.

The procedure for obtaining the 3 unknown parameters of the compound lens is then as follows :

  1. Draw a ray diagram for any 2 object points $O_1, O_2$ the same height above the optical axis. Construct the corresponding image points $I_1, I_2$ for the whole compound lens system using the method given above.

  2. Rays from $O_1, O_2$ are parallel to the axis so they are refracted at the same point on the back principal plane BPP and both pass through the back focal point $F_2$ before reaching $I_1, I_2$. Extend the line $I_1 I_2$. Where this line intersects the principal axis is the back focal point $F_2$ (BFP). Where this line intersects the line $O_1 O_2$ extended is the location of the back principal plane (BPP) which intersects the axis at pole $P_2$. The distance $P_2 F_2$ is the effective focal length $F$ of the compound lens.

Use the same procedure in reverse to find the position of the front principal plane :

  1. Construct any 3rd image point $I_3$ the same height as $I_1$ (or $I_2$) and on the same side of the axis. Construct the image(s) of $I_3$ formed by the lenses in reverse order $L_2, L_1$ - hence locate final image point $O_3$ corresponding to $I_3$.

  2. Extend line $O_1 O_3$. Where it intersects the axis is the position of the front focal point (FFP) $F_1$. Where it intersects the line $I_1 I_3$ is the location of the front principal plane, which intersects the axis at pole $P_1$. The distance $F_1 P_1$ is again the effective focal length $F$. It should be the same as found in step 2.

![enter image description here

To construct ray diagrams for the compound lens start with the known positions of front and rear lenses $L_1, L_2$. The positions of the front and rear principal planes $P_1, P_2$ of the compound lens are inserted relative to positions of $L_1, L_2$. Finally the focal points $F_1, F_2$ are positioned at the same distance $F$ in front of $P_1$ and at the back of $P_2$. In subsequent ray tracing the points $L_1, L_2$ are ignored.

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  • $\begingroup$ When you say two object points O1 and O2 "the same distance from principal axis" you mean the same height from the optical axis, i.e., the line where (in the picture) F1 and F2 are marked, right? $\endgroup$
    – Bidon
    Commented Mar 16, 2020 at 16:39
  • $\begingroup$ Yes that is correct. I am in the process of making some images so that the construction is clearer. $\endgroup$ Commented Mar 16, 2020 at 16:57
  • $\begingroup$ That would be incredible, I knew that it was a lot to ask, but that would help a lot $\endgroup$
    – Bidon
    Commented Mar 16, 2020 at 17:04
  • $\begingroup$ I hope the construction in steps 1 to 4 does give the right answers. I made up the method myself! Please let me know if you find it doesn't work. $\endgroup$ Commented Mar 16, 2020 at 18:37

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