Imagine we have two lens, one convex and one concave, spaced in such a way that the convex lens is before the concave lens. Now each lens has its own focus length and both are spaced such that the concave lens focus is to the right of the convex's. Furthermore, imagine that an arrow with a real length (in the $+y$-direction) is placed at the focus of the convex lens.

It seems that the rays that are incident on the convex lens will be redirected so as to emerge parallel to the optical axis. The image will then appear to be coming from infinite to the concave lens. Now these rays, once going through the concave lens, will diverge with the rays appearing to come from the concave lens focal point.

The question is: what is the magnification of the arrow by the optical system? In other words, can the arrow have a magnification and if so, how?

  • $\begingroup$ Hello kuantumbro. I didn't downvote it. But, the question has errors. First of all, why there are two cancaves at your 3rd line... $\endgroup$ Commented Oct 9, 2012 at 2:10
  • 3
    $\begingroup$ It's hard to be sure of what the situation is. A diagram would help. $\endgroup$
    – DarenW
    Commented Oct 9, 2012 at 5:35
  • 1
    $\begingroup$ It would have been nice to see a diagram, but two downvotes are unfair. It's a perfectly reasonable question. $\endgroup$ Commented Oct 9, 2012 at 6:43
  • $\begingroup$ And like both said, a diagram would be helpful @kuatumbro. You could provide somewhat good picture using Paint. It won't be too hard I think so. But, those two downvotes are not necessary..! (They're commentless either...) $\endgroup$ Commented Oct 9, 2012 at 7:37
  • $\begingroup$ Ok sorry about that, next time I'll be sure to include pertinent diagrams and such. Thanks for the tips though. $\endgroup$
    – kuantumbro
    Commented Oct 9, 2012 at 15:48

1 Answer 1


I suspect you're worrying about the appearance of infinities in the equations. For example the equation for the concave lens with the object at the focal point $f$ gives:

$$ \frac{1}{f} + \frac{1}{v} = \frac{1}{f} $$

and therefore $v$ = infinity. This leaves you wondering how to calculate the magnification for the second step. If you want to do this rigorously put the object at $f + \delta$, where $\delta$ is some small distance that you'll eventually set to zero. The equation for the first (concave) lens now gives:

$$ \frac{1}{f + \delta} + \frac{1}{v} = \frac{1}{f} $$


$$ \frac{1}{v} = \frac{1}{f} - \frac{1}{f + \delta} $$

The image from the concave lens become the object for the convex lens, so take this expression for $1/v$ and put it in the convex lens equation as $1/u$:

$$ \frac{1}{f} - \frac{1}{f + \delta} + \frac{1}{v} = \frac{1}{-F} $$

where $F$ is the focal length of the convex lens. A quick rearrangement gives:

$$ \frac{1}{v} = \frac{1}{f + \delta} - \frac{1}{f} - \frac{1}{F} $$

and now set $\delta$ to zero and we get:

$$ \frac{1}{v} = - \frac{1}{F} $$

so the image is a virtual image at a distance $F$, which is exactly what you'd expect since parallel rays form a virtual image at the focal point. The magnification is just the distance of the final (virtual) image divided by the distance of the original object, i.e. $F/f$.

  • $\begingroup$ This was exactly what I was worried about. Thank you for your insight. It still seams weird that instead of becoming a point it becomes magnified at the focal point. $\endgroup$
    – kuantumbro
    Commented Oct 9, 2012 at 15:46
  • 1
    $\begingroup$ @kuantumbro: you can look at it this way. The magnification at the convex lens is $\infty/f$ while the magnification at the concave lens is $F/\infty$. The overall magnification is the product of the two giving $F/f$. However you'd get your knuckles rapped for doing arithmetic with $\infty$ in class. Adding a distance $\delta$ that you'll later set to zero is a trick to get around the infinities. $\endgroup$ Commented Oct 9, 2012 at 15:51
  • 1
    $\begingroup$ @JohnRennie - that's the difference between physics and maths. In physics infinities cancel ! $\endgroup$ Commented Oct 9, 2012 at 16:25
  • $\begingroup$ Thanks, I'm going to ponder over this for a bit until I become convinced. $\endgroup$
    – kuantumbro
    Commented Oct 10, 2012 at 22:23

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.