I was working on my physics homework when I incorrectly copied part of a table for an optics problem with a two lens system, and I found myself trying to find the total magnification of a system of two converging lenses with the object at the focal length of the first lens. I found the final image distance by treating the first image distance as if it were infinity, leading the second image distance to be the second focal length. I'm not sure that's right in this case. The final part of the question was to calculate the total magnification, which gave me an expression that looked like


Did I do something wrong? What is the physical result of this setup?Diagram of the setup

  • $\begingroup$ I think your answer is correct. $\endgroup$
    – Allure
    Oct 22, 2019 at 23:42

2 Answers 2


For all of these specific, weird cases, I usually start by getting a general solution. I make the solution as direct as possible: derive the expressions that spit out the required outputs straight from the given inputs. Then I plug in the numbers for the weird case and see what happens.

For the first lens, image position and image size are given by:

$$ s_1' = - \frac{f_1 s_1}{f_1-s_1} $$ $$ h_1' = - \frac{h_1 s_1'}{s_1} $$

For the second lens at distance $d$, object position and size are:

$$s_2 = d - s_1' = d + \frac{f_1 s_1}{f_1-s_1}$$ $$h_2 = h_1' = - \frac{h_1 s_1'}{s_1}$$

So image position and size are:

$$ s_2' = - \frac{f_2 s_2}{f_2-s_2} = \frac{d f_{1} f_{2} - {\left(d - f_{1}\right)} f_{2} s_{1}}{d f_{1} - f_{1} f_{2} - {\left(d - f_{1} - f_{2}\right)} s_{1}} $$ $$ h_2' = - \frac{h_2 s_2'}{s_2} = -\frac{f_{1} f_{2} h_{1}}{d f_{1} - f_{1} f_{2} - {\left(d - f_{1} - f_{2}\right)} s_{1}} $$

Finally, total magnification is:

$$ m = \frac{h_2'}{h_1} = -\frac{f_{1} f_{2}}{d f_{1} - f_{1} f_{2} - {\left(d - f_{1} - f_{2}\right)} s_{1}} $$

The weird case you're interested in is when $s_1 = f_1$. Let's plug it into the equations:

$$ s_2' = f_2 $$ $$ h_2' = - \frac{f_2 h_1}{f_1} $$ $$ m = - \frac{f_2}{f_1} $$

So it would seem that the system should work without any weird infinities after all. Interesting.

Some notes

Technically, I used simplifications that can hide a division by zero. A more proper way of doing all this would be to keep the ugly expressions as they are, without canceling out potential zeros:

$$ s_2' = \frac{{\left(d + \frac{f_{1} s_{1}}{f_{1} - s_{1}}\right)} f_{2}}{d - f_{2} + \frac{f_{1} s_{1}}{f_{1} - s_{1}}}$$ $$ h_2' = -\frac{f_{1} f_{2} h_{1}}{{\left(d - f_{2} + \frac{f_{1} s_{1}}{f_{1} - s_{1}}\right)} {\left(f_{1} - s_{1}\right)}}$$ $$ m = -\frac{f_{1} f_{2}}{{\left(d - f_{2} + \frac{f_{1} s_{1}}{f_{1} - s_{1}}\right)} {\left(f_{1} - s_{1}\right)}}$$

Plugging in $s_1 = f_1$ directly will now properly lead to divisions by zero so we can't do it. What we can do, however, is calculate the limit of these expressions as $s_1$ approaches $f_1$, which leads to the exact same solution as before.

BTW, I used SageMath for all of this. I'd rather not do it by hand.

  • $\begingroup$ Another interesting thing: it doesn't matter how far away the lenses are anymore. This makes sense because any rays originating from the focal point will become parallel on the other side of the convex lens. So it doesn't matter how far away the second lens is, it will still catch the same parallel rays and do the inverse operation: direct them towards its point. $\endgroup$ Oct 23, 2019 at 6:13
  • $\begingroup$ That is interesting, thank you for your solution this is exactly what I was looking for. $\endgroup$
    – Max H.
    Oct 23, 2019 at 14:17

This problem can also be solved by using the fact that paraxial bundles of rays positioned at the focal point will emerge parallel. For rays that are offset from the focal point by a certain height, the rays will still emerge parallel but at an angle.

enter image description here

If the focal length of the first lens is $f_1$ and the object height is $y_o$, then the ray which passes through the center of the first lens will have an angle

$$\theta \approx y_0/f_1$$

with respect to the optical axis.

Since this ray passes through the lens without refraction, and all the other rays will be parallel to it, we can say that the rays emerge from the first lens all parallel to each other and at an angle of $\theta$ with respect to the optical axis. However, the same reasoning can be applied to the second lens. If we let the final image height be $y_i$, then we have

$$\theta \approx -y_i/f_2$$

(If you draw $y_o$ above the axis, you can reason physically that $y_i$ should be below the axis, so following the typical sign convention, $y_i$ would be negative, thus the negative sign to appropriately define $\theta$.)

Finally, we have

$$ \frac{y_i}{y_0}=-\frac{f_2}{f_1} $$

Here's my crude drawing:

enter image description here


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