# Longitudinal magnification

I want to prove that if an object is small in length and lying along the principal axis then

$$M = -\frac{dv}{du} = -\left(\frac{v}{u}\right)^2$$

where $$M$$ is the longitudinal magnification

In geometrical optics the following relation between the longitudinal positions of object and source (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

$$M = \frac{v^2}{u^2}$$

• JTS where you took derivative of (1/u)+(1/v) but derivative with respect to time should be taken – user221619 Apr 5 at 0:41
• @user221619 $\left (\frac{1}{u} + \frac{1}{v}\right)$ can be seen as a function of time if you let $u$ (and correspondingly $v$) depend on time, but I do not think it is important. The important thing is to notice that the variation of $v$ corresponding to the variation of $u$ must be such that the variation of $\left (\frac{1}{u} + \frac{1}{v}\right)$ is zero. For example you can take the derivative of $\left (\frac{1}{u} + \frac{1}{v}\right)$ with respect to $u$: $\frac{d \frac{1}{u} + \frac{1}{v}}{du} = -\frac{1}{u^2}\,\frac{du}{du} -\frac{1}{v^2}\,\frac{dv}{du} = 0$ and it works out. – JTS Apr 5 at 9:53