(a) A luminous point is moving at speed$V_0$toward a spherical mirror with a radius of curvature$\mathit{r}$, along the central axis of the mirror. Show that the image of this point is moving at the speed

${\mathbf{v}}_{\mathbf{I}}\mathbf{=}\mathbf{-}{\left(\frac{r}{2p-r}\right)}^{\mathbf{2}}{\mathbf{v}}_{\mathbf{0}}$

Where,$\mathit{p}$ is the distance of the luminous point from the mirror at any given time. Now assume the mirror is concave, with$\mathit{r}\mathbf{=}\mathbf{15}\mathbf{}\mathit{c}\mathit{m}$.and let${\mathit{V}}_{\mathbf{0}}\mathbf{=}\mathbf{5}\mathbf{}\mathit{c}\mathbf{m}\mathbf{/}\mathbf{s}$. Find${\mathit{V}}_{\mathbf{1}}$when (b)$\mathbf{p}\mathbf{=}\mathbf{30}\mathbf{}\mathit{c}\mathit{m}$(far outside the focal point), (c) $\mathbf{p}\mathbf{=}\mathbf{8}\mathbf{.}\mathbf{0}\mathbf{}\mathit{c}\mathit{m}$(just outside the focal point), and (d)$\mathit{p}\mathbf{=}\mathbf{10}\mathbf{}\mathit{m}\mathit{m}$(very near the mirror).

For a given luminous object to be placed in front of a concave mirror, the image distance can be given using the lens equation. If the object is placed at infinity, then the image is formed at the focal point of the mirror, but if the object is within the focal point of the mirror, then the object formed is virtual. Now, the object when placed after the focal point before the radius of curvature, the image distance is far from the mirror, while for the object at a distance after the radius of curvature, the image distance is near to the mirror. Again, you know that the speed of an object is defined by its distance within the travel time. Thus, using this concept, you calculate the speed values.

Formulae:

The mirror equation is,

$\frac{1}{f}=\frac{1}{i}+\frac{1}{p}$ ….. (i)

Where, $p$is the pole, $f$is the focal length, and $i$is the image distance.

The speed of the object is,

$v=\frac{dx}{dt}$ ….. (ii)

Where,$dx$ is the distance travelled by the object and$dt$ is the time taken to travel.

Using equation (i), the formula of image distance can be given as:

$i=\frac{pf}{p-f}$ ….. (iii)

But, you know that the focal length is half the radius$f$ curvature. That is given as:

$f=\frac{r}{2}$

Substituting the above value in equation (iii), you get the value of image distance as:

$$\begin{gathered} i=\frac{p\frac{r}{2}}{p-\frac{r}{2}} \\ =\frac{pr}{2p-r} \end{gathered}$$

Now, the speed of the light is given using equation (ii) as follows:

$$\begin{gathered} v_I=\frac{di}{dt} \\ =\frac{d}{dt}\left(\frac{pr}{2p-r}\right) \\ =\frac{\left(2p-r\right)\frac{d}{dt}\left(pr\right)-pr\frac{d}{dt}\left(2p-r\right)}{{\left(2p-r\right)}^2} \end{gathered}$$

Since, the speed of the object can be given using equation (ii) as follows:

$V_0=-\raisebox{1ex}{$dp$}\!\left/ \!\raisebox{-1ex}{$dt$}\right.$

Thus, the speed of the luminous light can be given as:

$v_I=-{\left(\frac{r}{2p-r}\right)}^2{v}_0$

Hence, the image is moving at the speed $v_I=-{\left(\frac{r}{2p-r}\right)}^2{v}_0$is shown in the part of the calculation.