[Physics] Where does the last term come from in the two-lens formula: $\frac{1}{f}=\frac{1}{f_1} +\frac{1}{f_2} -\frac{d}{f_1f_2}$

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I can derive the formula for lenses in contact: i.e, $$\frac{1}{f}=\frac{1}{f_1} +\frac{1}{f_2}.\tag{1}$$

But for two lenses separated by a distance $d$ I can't seem to get:

$$\frac{1}{f}=\frac{1}{f_1} +\frac{1}{f_2} -\frac{d}{f_1f_2}.\tag{2}$$

For the first derivation I let the image distance for the first lens $v_1$ equal the negative object distance for the second lens. ie. $v_1=-u_2$

When trying to derive the second formula I let: $u_2=d-v_1$ , and then followed the same procedure. I ended up with a very tedious algebraic expression. Did I go wrong with $u_2=d-v_1$? How do you derive the second formula?

Best Answer

I can't resist to show how this is done using ray transfer matrices. There are two key parameter at any point along a light ray: the distance $x$ of the point from the optical axis, and the angle $\theta$ of the ray with the horizontal. Then any optical component of the system can be represented as a $2\times 2$ matrix which transform the pair $(x,\theta)$ for the incoming ray into the pair $(x',\theta')$ for the outgoing ray:

$$\begin{pmatrix}x'\\\theta'\end{pmatrix}=\begin{pmatrix}\times&\times\\\times&\times\end{pmatrix}\begin{pmatrix}x\\\theta\end{pmatrix}.$$

By using what you know, you can easily write down that

  • a thin lens has a matrix $$L=\begin{pmatrix}1&0\\-\dfrac{1}{f}&1\end{pmatrix}$$ where $f$ is the focal distance;
  • a thickness $d$ of empty space has a matrix $$S=\begin{pmatrix}1 & d\\0&1\end{pmatrix}.$$

Onto your problem: we have a lens of focal $f_1$ (matrix $L_1$), empty space and another lens of focal $f_2$ (matrix $L_2$), so the matrix for the whole system is simply the product of the matrices, in reverse order

$$M=L_2\ S\ L_1 =\begin{pmatrix}1&0\\-\dfrac{1}{f_2}&1\end{pmatrix} \begin{pmatrix}1 & d\\0&1\end{pmatrix} \begin{pmatrix}1&0\\-\dfrac{1}{f_1}&1\end{pmatrix} =\begin{pmatrix}1-\dfrac{d}{f_1}&d\\-\left(\dfrac{1}{f_1}+\dfrac{1}{f_2}-\dfrac{d}{f_1f_2}\right)&1-\dfrac{d}{f_2}\end{pmatrix}$$

and you can read the focal distance in the bottom left corner! As you see, 99% of my exposition is just explaining the method. The actual computation is that trivial, systematic, triple matrix product. No monkeying about with geometry all the time: you just need to do it once to deduce the matrices for common components.

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