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.
[Physics] Proof of minimum distance between object and image of a convex lens image being 4*focal length
homework-and-exerciseslensesoptics
Best Answer
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:
$$ v = \frac{uf}{u - f} $$
so:
$$\begin{align} s &= u + \frac{uf}{u - f} \\ &= u \frac{u-f}{u-f} + \frac{uf}{u - f} \\ &= \frac{u(u-f) + uf}{u - f} \\ &= \frac{u^2}{u - f} \end{align}$$
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:
$$ g(u) = u^2 $$
$$ dg/du = 2u $$
And:
$$ h(u) = u - f $$
$$ dh/du = 1 $$
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 $u = 0$.