A point object $P$ moves towards a convex mirror with a constant speed $v$ , along its optic axis. The speed of the image
A. Is always $ < v$
B. May be $ > , = or < v $ depending on the position of $P$
C. Increases as $P$ comes closer to the mirror
D. Decreases as $P$ comes closer to the mirror

Hint: For the convex mirror, as the object travels from infinity towards the pole, the image begins to move from focus to pole. Within the same time interval, the image always covers less distance than the object. We can also mathematically find the answer to this question by using the mirror formula.

Formula used:
$ \frac{1}{x_{i}}+\frac{1}{x_{o}}=\frac{1}{f}$




Complete answer:
The mirror equation can be written as follows:
$\frac{1}{x_{i}}+\frac{1}{x_{o}}=\frac{1}{f}$ .............(1)

where $x_{i}$ and $x_{o}$ are distances of the image and the object.
We need to differentiate the equation (1) w.r.t ‘t’.

$-\frac{1}{x_{i}^{2}}\frac{dx_{i}}{dt}-\frac{1}{x_{o}^{2}}\frac{dx_{o}}{dt}=0$ ………….(2)

In the above equation, we can say that $\frac{dx_{i}}{dt}$ is the speed/velocity of the image $v_{i}$ whereas, $\frac{dx_{i}}{dt}$ is the speed/velocity of the object. Therefore, we can write equation (2) as follows:

$-\frac{v_{i}}{x_{i}^{2}}-\frac{v_{o}}{x_{o}^{2}}=0$

or, ${v_{i}}=-\frac{x_{i}^{2}}{x_{o}^{2}}\lgroup{v_{o}}\rgroup$ ................(3)

Let’s rewrite equation (1) in the following matter,
$x_{i}=\frac{fx_{o}}{x_{o}-f}$
Squaring the above equation, we get
$x_{i}^{2}=\frac{f^{2}x_{o}^{2}}{\lgroup~x_{o}-f~\rgroup^{2}}$
Substituting the value $x_{i}^{2}$ in equation (3), we have
$v_{i}=\frac{f^{2}}{\lgroup~x_{o}-f~\rgroup^{2}}\lgroup~v_{o}\rgroup$
Therefore, from the above equation, it can be said that the speed of the image is less than the object's speed.
The correct option is A.




Note: Usually, in the mirror formula v is taken as the image distance while u is taken as the object distance. But in this question, v is mentioned as the object’s speed. Be careful in this part and don’t get confused. Also, it was assumed that the same mirror (convex) has been used in the entire problem making focal length (f) constant.