Velocity of image Suppose I have an object placed in front of a convex lens at a distance twice its focal length. Then an image is also obtained at twice its focal length to the right of the lens. Suppose it is not a point object, say it has a height of $x$ cm and it starts moving towards the lens with a speed of $y$ cm/s. 
It is understandable that the image too moves along the principal axis @$y$ cm/s. But here the size of image also keeps on increasing, thus acquiring a kind of vertical velocity. 
Do we consider it in our velocity of image? I can't really understand why we should, or at least how, as vertical velocity might not be same for all points lying on the image itself. If we just think of magnification increase, then we would be getting vertical velocity only for the tip of image, not thw image as a whole I guess? 
Please someone to help me out in this. 
 A: The basic relationships between the position of a point on the object-side and its 3D image are given by:
$$ \frac{1}{f} = \frac{1}{s} + \frac{1}{s'} $$
$$ \frac{h'}{h} = -\frac{s'}{s} $$
Here, $f$ is the lens' focal distance, $s$ is the distance of the point from the lens on the object-side, $s'$ is the distance of the points' image on the image-side, and $h$ and $h'$ are the vertical distance of the point and its image from the central axis.
1) Moving the sensor to track the image
For scenario 1, we keep the lens fixed and keep the sensor at the exact position $s'$ as the point moves. This way, the 2D image we get on the sensor will always be sharp. By just using the equations above, we can derive the exact sensor position $s'$ and the height of the image $h'$.
$$ s' = \frac{fs}{s-f} $$
$$ h' = -\frac{fh}{s-f} $$
As the object moves towards the lens at speed $ds/dt$, at the constant height $h$, we can calculate the sensor's speed $ds'/dt$ and the vertical speed of the point on the image $dh'/dt$:
$$ \frac{ds'}{dt} = - \frac{f^2}{\left( s-f \right)^2} \frac{ds}{dt} $$
$$ \frac{dh'}{dt} = \frac{fh}{\left( s-f \right)^2} \frac{ds}{dt} $$
2) Static imaging system, no refocusing
Things get a bit more complicated if neither the lens or the sensor move. Then the camera is focused for just one object distance $s$, and the sensor is at position $s'$.
For a sharp object at the distance $s$, we can calculate the focused magnification $m$ (which is negative, as the image is always upside-down):
$$ m = \frac{h'}{h} = - \frac{f}{s - f} $$
As the object moves to a defocused position $s_d$, the magnification will change and I'll call this defocused magnification $m_d$:
$$ m_d = \frac{h'_d}{h} = - \frac{bf}{s_d-s_{ep}} $$
Since we're dealing with defocused points, the lens' entrance pupil becomes important. In this equation, it is manifested as the bellows factor $b$ and the entrance pupil position $s_{ep}$. Those can be calculated from pupil magnification $p$, which can be assumed to be equal to 1 for symmetric lens assemblies:
$$ b = 1 - \frac{m}{p} $$
$$ s_{ep} = f \left( 1 - \frac{1}{p} \right) $$
By rearranging the equation for defocused magnification $m_d$, we can calculate the height of the image point $h_d'$ and then differentiate it to get its speed as the defocused object moves at speed $ds_d/dt$:
$$ h'_d = \frac{-bfh}{s_d-s_{ep}} $$
$$ \frac{dh'_d}{dt} = \frac{bfh}{\left( s_d - s_{ep} \right) ^2} \frac{ds_d}{dt} $$
3) Block-focusing: moving the lens to keep the object in focus
The simplest focusing mechanism used in cameras works by moving the whole lens assembly back and forth while keeping the sensor stationary. This is called block focusing.
Since the lens is moving, but sensor isn't, I'll introduce $x$ as the distance of the object from the sensor. $s$ and $s'$ are still distances from the object to the lens and from the lens to the sensor, but they will be changing now to keep the 2D image on the sensor sharp.
For an infinitely thin lens, this distance $x$ would equal $s + s'$, but real lens have a certain thickness $w$. This isn't the actual thickness of the glass, but the distance between image- and object-side principal planes (this distance can be negative).
$$ x = s + s' + w $$
By substituting $s = f s' / (s' - f)$ (from the first equation in the article) and solving the quadratic equation, we can calculate where to put the lens relative to the sensor:
$$ s'_{1,2} = \frac{1}{2} \left( \pm \sqrt{w-x} \sqrt{4f+w-x} - w + x \right) $$
(This can probably be expressed more tidily, but I just used Wolfram Alpha. The main thing to notice is that there are two sharp configurations $s'_1$ and $s'_2$ for every object-to-sensor distance, because optical systems are symmetrical. By just swapping the object and sensor positions, you can turn a system with magnification 2 into a system with magnification 0.5.)
By differentiating this, we get the required lens' speed as object moves at speed $dx/dt$:
$$ \frac{ds'_{1,2}}{dt} = \mp \frac{\left( \sqrt{w-x}\mp\sqrt{4f-x+w}\right)^2}{4\sqrt{w-x}\sqrt{4f-x+w}} \frac{dx}{dt} $$
(Note the $\mp$ as opposed to $\pm$ when matching $s'_1$ and $s'_2$ to their derivatives.)
By using the first two equations from the article, we can derive:
$$ h_{1,2}' = h - \frac{h}{f} s'_{1,2} $$
This is the vertical position of the 2D image. Differentiating this gets us the speed:
$$ \frac{dh'_{1,2}}{dt} = - \frac{h}{f} \frac{ds'_{1,2}}{dt} $$
4) Internal focusing
I should mention what is probably the most widespread modern camera focusing mechanism: internal focusing. It works not by moving the entire lens assembly, but only some internal components.
This makes both the lens' focal distance $f$ and effective thickness $w$ variable. Since both will now have their time derivatives, things start to get complicated to model and I'm not going to attempt it here.
