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

like in the text in the image


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

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

  • $\begingroup$ JTS where you took derivative of (1/u)+(1/v) but derivative with respect to time should be taken $\endgroup$ – user221619 Apr 5 at 0:41
  • $\begingroup$ @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. $\endgroup$ – JTS Apr 5 at 9:53

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