A point object O is located on the principal axis at a distance of 20 cm from pole P of a concave surface of radius of curvature 10 cm. The object O is given velocity ($4 \hat i+3 \hat j$) cm/s as shown. Then the velocity of image, when the object is at O, is:
$\text{A.}\quad (8\hat i + 9\hat j ) cm/s$
$\text{B.}\quad (24\hat i + 9\hat j ) cm/s$
$\text{C.}\quad (8\hat i - 9\hat j ) cm/s$
$\text{D.}\quad (-24\hat i + 8\hat j ) cm/s$

Hint: When light gets refracted from a general spherical surface made of some different refractive index, that case comes under the category of refraction due to spherical surface. We have to be careful about the fact that rays are travelling from rarer to denser medium or from denser to rarer medium.
Formula used:
$\dfrac{n_2}{v} - \dfrac{n_1}{u} = \dfrac{n_2-n_1}{R},\ \dfrac{h_i}{h_o} = \dfrac{n_1v}{n_2u}$

Complete answer:
Given: $n_2 = 1,\ n_1 = 1.5,\ R = -10cm,\ u=-20$
Using $\dfrac{n_2}{v} - \dfrac{n_1}{u} = \dfrac{n_2-n_1}{R}$, we get
$\dfrac{1}{v} - \dfrac{1.5}{-20} = \dfrac{1-1.5}{-10}$
$\dfrac{1}{v} = -\dfrac{0.5}{20}$
$\implies v = -40cm$
Now, x-component velocity of image may be defined as the rate of change of image distance from pole. This distance could be calculated by using $\dfrac{n_2}{v} - \dfrac{n_1}{u} = \dfrac{n_2-n_1}{R}$.
Rate of change of image distance could be calculated by differentiating this expression:
$-\dfrac{ n_2 }{v^2}\dfrac{dv}{dt} + \dfrac{ n_1 }{u^2}\dfrac{du}{dt} = 0$
Or $\dfrac{dv}{dt} = v_x = \dfrac{n_1}{n_2} \left(\dfrac{v}{u}\right)^2u_x$
Also velocity of image = $4 \hat i+3 \hat j$cm/s
Thus, $u_x = 4 cm/s\ and \ u_y = 3 cm/s$
Putting values in $\dfrac{dv}{dt} = v_x = \dfrac{n_1}{n_2} \left(\dfrac{v}{u}\right)^2u_x$, we get;
$v_x = \dfrac{1.5}{1} \left(\dfrac{-40}{-20}\right)^24 = 24cm/s$
Now, the vertical component of the image velocity could be calculated by using the rate of change of image height which is $h_i$.
Thus, using $\dfrac{h_i}{h_o} = \dfrac{n_1v}{n_2u}$
$\implies h_i = \left(\dfrac{n_1v}{n_2u} \right) h_o$
$v_y = \dfrac{dh_i}{dt} = \left( \dfrac{n_1v}{n_2u}\right) \dfrac{dh_o}{dt}$
Now, putting the values:
$v_y = \left( \dfrac{1.5(-40)}{1(-20)}\right) \times 3$ [as $\dfrac{dh_o}{dt} = 3cm/s$]
$\implies v_y = 9 cm/s$

So, the correct answer is “Option B”.

Note:
One can easily mark the correct option by knowing velocity in any one direction. The whole process is explained so that a student can practice the authentic process of finding the velocity of an image. Students should not remember these formulas, rather should practice the formulas by differentiation. Also the height of the object is not actually the length of the object above the x-axis, but the distance of it above the principal axis.