I have found the formula for the effective focal length $f$ of two thin lenses with focal lengths $f_1$ and $f_2$ separated by distance $d$ to be $$ \frac 1f=\frac 1{f_1}+\frac 1{f_2}-\frac d{f_1f_2}. $$ However, I can't seem to find how $f$ is defined. Is it the distance from the first lens to the final focal point or the distance from the second lens to the final focal point? Or neither?

  • $\begingroup$ The distance for the lens or lens set to converge an incoming parallel beam to a single point. For a diverging lens or set extend the diverging rays backwards until they converge on the optic axis. The distance from this point to the first deflection from parallel is the focal length, usually expressed as -ve for a divergence. $\endgroup$ – Arif Burhan May 18 '16 at 19:47

It is the distance from the image plane to the rear principal plane. You can find the location of this plane by projecting the image ray backwards through the system to where it crosses the projection of the object ray. This is sometimes also referred to as the effective focal length (EFL) of the system, and is true for both simple as well as complicated systems. The distance from the rear lens to the image plane is simply the back focal distance (BFD). The difference between the EFL and BFD can be found by the formula:

delta = -(d/n)*(f/f1) = -(EFL-BFD); where n=1 in air

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  • $\begingroup$ If I add the formula you gave for delta to the formula for BFD, f2*(f1-d)/(f1+f2-d), it does not appear to equal the formula for effective focal length. I derived delta myself and arrived at the same formula you gave so I assume it's correct, but I can't figure out why adding delta to BFD doesn't yield the formula for EFL that I have found in numerous sources. Do you know where the discrepancy lies? $\endgroup$ – David Webb Apr 11 '16 at 3:02
  • $\begingroup$ Your formula derives just fine. Take your time, sketch it all out, make sure you have the definitions correct, and post your work. I'll check it for errors if you'd like, but you have everything correct so far. $\endgroup$ – Andrew Stockham Apr 12 '16 at 4:26
  • $\begingroup$ I uploaded a photo of my work to photobucket: i76.photobucket.com/albums/j23/dwebb1211/work_zpsqd05h4ej.png Please let me know if anything is wrong. $\endgroup$ – David Webb Apr 13 '16 at 17:02
  • $\begingroup$ You have the formula for delta as -df2/f1, it is -dEFL/f1. $\endgroup$ – Andrew Stockham Apr 13 '16 at 17:10

The focal point for the combined combination of lenses is a distance f from the secondary principal point of the second lens. If you are approximating the lenses as thin then the answer you're looking for is the distance from the second lens to the final focal point.



  • $\begingroup$ I might be missing something, but I believe the reference you cite defines the distance from the second lens to the focal point as s2'', not f. This appears to have a different formula than f. $\endgroup$ – David Webb Apr 5 '16 at 22:00
  • $\begingroup$ You're right - s2" is the back focal length - as you say a different formula... $\endgroup$ – M. Enns Apr 6 '16 at 0:17
  • $\begingroup$ OK, what about page 3 of this document - It shows the distance from the principal plane of the second lens to F' as the combined focal lenght and, while the don't spell it out explicitly, it sure looks like F' is the focal point of the combined lenses. home.uni-leipzig.de/prakphys/pdf/VersucheIPSP/Optics/… $\endgroup$ – M. Enns Apr 6 '16 at 0:44
  • 1
    $\begingroup$ A system of two thin lenses can be treated as single equivalent thick lens. Thick lenses have 2 principal planes. f is the distance from the back principal plane to the focal point for rays going from left to right or the distance from the front PP to the focal point for rays going from right to left. So, it looks like the answer to your original question is f is the distance from the focal point to either lens depending on which way the rays are going. $\endgroup$ – M. Enns Apr 6 '16 at 3:04

protected by Qmechanic Sep 16 '16 at 15:14

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