Geometrical optics. Photometry. Quantum nature of light > Optical systems > Problem 13.3.9

Statement

$13.3.9.$ The subject is moving towards the movie camera at a speed of $v$. How fast do you need to change the focal length of the lens and the depth of the movie camera so that the image size remains the same if the magnification given by the movie camera is $k$?

Solution

Let's define "magnification" as
$k = \frac{s'}{s}$ (1)
where $s'$ is the image distance from len and $s$ is the object distance from camera.
The rate of change for the object-to-len's distance is
$\dot{s} = -v$ (2)
it has negative sign due to the object goes towards the camera.
The relation with focal distance is given by
$\frac{1}{s}+\frac{1}{s'} = \frac{1}{f}$ (3)
Differentiating (3) respect to time,
$\frac{\dot{s}}{s^2}+\frac{\dot{s'}}{s'^2} = \frac{\dot{f}}{f^2}$ (4)
From (1),
$\dot{s'} = k\dot{s}$ (5)
Putting (2) and (5) into (4):
$\frac{\dot{f}}{f^2} = -\frac{v}{s^2}\left(1+\frac{1}{k}\right)$ (6)
From (3), and using (1),
$\frac{1}{f} = \frac{1}{s}\left(1+\frac{1}{k}\right)$ (7)
Substituting (7) into (6),
$\dot{f} = \frac{df}{dt} = \frac{-vk}{k+1}$
the minus sign means that camera must reduce the focal distance at that rate.

Answer

$|\dot{f}| =\frac{vk}{k+1}$