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My textbook claims that for a pair of thin lenses separated by a distance $d$ the combined focal length of the system is:

$$\frac{1}{f}= \frac{1}{f_1} + \frac{1}{f_2} - \frac{d}{f_1f_2}.$$

Can this be shown using ray-tracing? If so, how? I find it interesting that there doesn't seem to be any dependence on whether the lenses are positive/negative or whether d is greater or smaller than the focal lengths of the particular lenses.

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  • $\begingroup$ Keep in mind that there is a sign convention for the focal length: diverging lenses have $f < 0$. So swapping one lens from converging to diverging changes the signs of two of the three terms on the right-hand side resulting in very different behavior. $\endgroup$ – dmckee Jun 11 '17 at 15:58
  • $\begingroup$ have you attempted to draw a ray diagram yourself? $\endgroup$ – JMLCarter Jun 11 '17 at 17:25
  • $\begingroup$ Yes, but I can't figure out how the rays behave in a general case. For example, if you have two lenses separated by the sum of their focal lengths, a parallel pencil of rays will come out of the system parallel again, which matches what the formula says (infinite focal length). But when I draw the lenses at an arbitrary distance from one another, all I can say is that the first one refracts the rays, say, inwards (it it's positive) and the second one refracts them even more and the crossing point with the optical axis defines the focal length. But I didn't manage to quantify it. $\endgroup$ – Piotr Jun 11 '17 at 17:36
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I haven't actually done the derivation but the approach you would take would be to write a ray tracing matrix for the whole system, including the object distance $s_1$ and the image distance $s_2$:

$$ \begin{bmatrix}x_f \\ \theta_f\end{bmatrix} = \begin{bmatrix}1 & s_2 \\ 0 & 1\end{bmatrix} \begin{bmatrix}1 & 0 \\ -1/f_2 & 1\end{bmatrix} \begin{bmatrix}1 & d \\ 0 & 1\end{bmatrix} \begin{bmatrix}1 & 0 \\ -1/f_1 & 1\end{bmatrix} \begin{bmatrix}1 & s_1 \\ 0 & 1\end{bmatrix} \begin{bmatrix}x_i \\ \theta_i\end{bmatrix} $$

Then you do the tedious matrix multiplication so that you get coefficients $A,B,C,D$ for your equation system in terms of $d,f_1,f_2,s_1,s_2$:

$$ \begin{bmatrix}x_f \\ \theta_f\end{bmatrix} = \begin{bmatrix}A & B \\ C & D\end{bmatrix} \begin{bmatrix}x_i \\ \theta_i\end{bmatrix} $$

When an image is formed, all the rays starting from position $x_i$ end up at $x_f$ regardless of their initial angle $\theta_i$. So in the equation $x_f = Ax_i + B\theta_i$, you can set $B=0$ and from there derive an expression for $\frac{1}{s_1} + \frac{1}{s_2}$ which is the focal length of the whole system.

This expression, which is hopefully the same as what your book says, will certainly depend on $f_1$, $f_2$, and $d$. If you plot each one while keeping the other two constant, you can see how they depend when e.g. one lens is negative and the other positive, or the distance is greater or smaller than the focal length.

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  • $\begingroup$ You don't really need the two outer matrices: the three matrix focus / freespace / focus sandwich in the middle will do; then element (2,1) in the product (C in your notation) is the negative reciprocal of the "focal length" (quotes because this focal length is the distance between the output plane and the focal plane, which is not the distance from principal plane to focal plane unless the lenses are thin). $\endgroup$ – WetSavannaAnimal Jun 12 '17 at 10:36

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