Longitudinal magnification

In geometrical optics the following relation between the longitudinal positions of object and image (respectively $u$ and $v$) together with the focal length $f$ is valid:

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

If the object is small and it has one of its ends at $u_1$, with the corresponding image at $v_1$, we can calculate the position of the image of the other end, $v_2$, in an approximate way using derivatives:

$$v_2 approx v_1 + frac{dv}{du} (u_2 - u_1)$$.

where $frac{dv}{du}$ is the derivative calculated at $u_1$.

The longitudinal magnification is the ratio between the length of the image and the length of the object:

$$ M = left|frac{v_2 - v_1}{u_2 - u_1}right|, $$

and using the approximate equation for the position in terms of the derivatives that we wrote above we see that

$$M approx left| frac{dv}{du} right|.$$

An easy way to calculate the derivative is considering that the variation of the quantity $frac{1}{u} + frac{1}{v}$ is zero for any variation of the position of the object $u$ and the corresponding variation of the position of the image $v$. So:

$$d left (frac{1}{u} + frac{1}{v}right) = 0$$.

We can express the variation using the variations of $u$ ($du$) and $v$ ($dv$) as

$$d left (frac{1}{u} + frac{1}{v}right) = -frac{1}{u^2},du -frac{1}{v^2},dv$$

still equal to zero. From $-frac{1}{u^2},du -frac{1}{v^2},dv = 0$ we obtain the expression for $frac{dv}{du}$:

$$frac{dv}{du} = - frac{v^2}{u^2}$$

The longitudinal magnification is then

$$M_{mathrm{long.}} = frac{v^2}{u^2}$$

Since the transverse magnification is

$$M_{mathrm{transv.}} = frac{v}{u}$$

then

$$M_{mathrm{long.}} = M_{mathrm{transv.}}^2$$