I can prove that the minimum distance between an object and its image, through a convex lens is 4*focal length, if I assume that the distance between the object and the lens is the same as the distance between the image and the lens, i.e. if: $$1/u + 1/v = 1/f$$ that $u=v$, but is there a way of showing that the distance $u+v$ is minimised at $u=v$, because I can't work one out.

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    $\begingroup$ Write down an equation for the quantity $s = u + v$ in terms of $u$. You can do this by starting with $s = u + v$ and substituting for $v$ using the equation you quote in your question. Now you find the value of $u$ for which $s$ is a minimum by setting $ds/du = 0$. The algebra is mildly tedious but not hideously difficult. You should find that $ds/du = 0$ when $u = 2f$. $\endgroup$ – John Rennie Jan 31 '14 at 15:59
  • $\begingroup$ @JohnRennie perhaps that should be an answer? (I'm not sure, but I suspect so) $\endgroup$ – David Z Jan 31 '14 at 18:18
  • $\begingroup$ @DavidZ: I must admit I have voted to close on the grounds it's homework. However if you feel it isn't I'll withdraw my VTC and post the answer. $\endgroup$ – John Rennie Jan 31 '14 at 18:22
  • $\begingroup$ @JohnRennie Well like I said, I'm not sure about this one. It's a homework-like question, sure, but AlexLipp does seem to identify a specific concept which is giving trouble, namely showing that $u=v$ minimizes $u+v$ with the given constraint. I don't feel qualified to judge this one on my own, and I wouldn't recommend you retract your close vote just because of me. I'd like to see what other people say about it. $\endgroup$ – David Z Jan 31 '14 at 18:25
  • $\begingroup$ @DavidZ: oh go on then :-) $\endgroup$ – John Rennie Jan 31 '14 at 19:29

Although the other answers are good, I will add my slightly different one. Rearranging the image equation gives; $$ u+v=\frac{1}{f}uv $$ On the left hand side you have the perimeter of a rectangle (well half of the perimeter, but it doesn't matter for the proof), and on the right hand side you have the area of the rectangle. It is a reasonably easy proof in geometry to show that a rectangle of fixed perimeter has minimum area when it is a square, i.e. when $u=v$. There are a couple of proofs of this fact here at math.SE.

Hence, the minimum of $u+v$ occurs when $u=v$. QED :)


If you define the function:

$$ s = u + v $$

then what you're trying to do is find out where $s$ is a minimum. Let's write down this function and see what it looks like. The first step is to substitute for either $u$ or $v$ so that $s$ is a function of only one variable. I'm going to substitiute for $v$. You give the equation:

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

and we can rearrange this to get:

$$ v = \frac{1}{\frac{1}{f} - \frac{1}{u}} $$

This looks ugly, so I'm going to multiply the top and bottom of the fraction on the right hand side by $uf$ to remove the fractions in the denominator. This gives the nicer equation:

$$ v = \frac{uf}{u - f} $$


$$\begin{align} s &= u + \frac{uf}{u - f} \\ &= u \frac{u-f}{u-f} + \frac{uf}{u - f} \\ &= \frac{u(u-f) + uf}{u - f} \\ &= \frac{u^2}{u - f} \end{align}$$

I quite like to sketch up a graph to check the function I'm working with looks sensible. Let's choose our units of length to make $f = 1$ and graph $s(u)$ (with a bit of help from Microsoft Excel):


This all looks sensible. It reaches a minimum around $u = 2$, which is actually $u = 2f$ because we're using units in which $f = 1$. The value of $u + v$ at the minimum is around $4f$ that would make $v = 2f$ as well. But to prove the minimum is at $u = 2f$ we need to calculate the the value of $ds/du$ and the minimum will be where the differential is zero i.e.

$$ \frac{ds}{du} = 0 $$

Calculating the derivative is straightforward but messy. I'll go through all the steps but feel free to skip over them if I'm going too slow. First we need the rule for differentiating a fraction - you'll find this proved in any elementary book on calculus:

$$ \frac{d}{dx} \frac{g}{h} = \frac{h\frac{dg}{dx} - g\frac{dh}{dx}}{h^2} $$

In our case:

$$ g(u) = u^2 $$

$$ dg/du = 2u $$


$$ h(u) = u - f $$

$$ dh/du = 1 $$

If we feed these results into the expression for $ds/du$ we get:

$$ \frac{ds}{du} = \frac{2u(u-f) - u^2}{(u - f)^2} $$

Which simplifies a bit to:

$$ \frac{ds}{du} = \frac{u(u - 2f)}{(u - f)^2} $$

And our condition for the minimum is that this is equal to zero, so we end up with:

$$ \frac{u(u - 2f)}{(u - f)^2} = 0 $$

And the only values for $u$ that satisfy this are $u = 0$ and $u = 2f$. Leaving aside the $u = 0$ solution for the moment, we find our function $s = u + v$ is minimised at $u = 2f$ and this is the result we need. If $u = 2f$ then the lens equation gives us $v = 2f$ and therefore the minimum value for $u + v = 4f$, which is what we set out to prove.

But hang on a sec, what about that other root at $u = 0$. This is obviously unphysical, but what does it mean? Well if you bring the object closer to the lens than the focal length you get a virtual image i.e. $v < 0$. Let's see what our function $s(u)$ looks like for $0 \lt x \lt f$:


What happens is that as we decrease $u$ towards $f$ the value of $u + v$ goes to $\infty$ then the other side of $u = f$ it switches to $-\infty$. Then as we decrease $u$ from $f$ towards $0$ the value of $u+v$ increases towards zero and comes to a maximum value of zero at $u = 0$. Setting $ds/du$ to zero found this maximum as well as the minimum at $u = 2f$.

So somewhat unexpectedly, we find that if you allow virtual images then the minimum value for $u + v$ is actually $-\infty$ and happens when $u = f$. If you take the more sensible view that you want the modulus of $u + v$ i.e. $s = |u + v|$, then the minimum value is zero at $u = 0$.


I think John Rennie's answer is wonderfully clear and detailed, but there's also a simple hackish (more intuitive?) way of approaching this problem.

Note that for a convex lens, the image and the object are "symmetric" in that you can swap their positions (and orientations). This is due to the principle of reversibility of the path of light, and is true more generally.

Since the distance betwen the image and the object is symmetric in $u$ and $v$, it will attain a local maximum or minimum when $u = v$ which happens only when $u = v = 2f \implies u + v = 4f$.

Since an object position must give a unique image position (and vice versa, due to reversibility of the path of light), changing $u$ better change $v$ monotonically. Otherwise, the function $v(u)$ will be multivalued (or equivalently the inverse $u(v)$ will not be well defined). So the only "special" point will be $u = v$ and it is infact a global maximum or minimum.

Deciding whether this is a minimum or maximum should be simple. Compare to any other pair $(u_1,v_1)$. If $U_1 + v_1 > 4f$ then $(u = 2f, v=2f)$ is indeed the minimum. For this test, we can let $u \rightarrow \infty , v \rightarrow f$. QED.


protected by Qmechanic Jul 26 '16 at 19:22

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