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(D) The mirror formula is given by $\frac{1}{f} = \frac{1}{v} + \frac{1}{u}$.For a concave mirror,$f = -12 \ cm$. The object distance is $u = -20 \ cm

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$9$ towards the mirror

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(D) The mirror formula is given by $\frac{1}{f} = \frac{1}{v} + \frac{1}{u}$.For a concave mirror,$f = -12 \ cm$. The object distance is $u = -20 \ cm$.Using the mirror formula: $\frac{1}{-12} = \frac{1}{v} + \frac{1}{-20} \implies \frac{1}{v} = \frac{1}{20} - \frac{1}{12} = \frac{3-5}{60} = -\frac{2}{60} = -\frac{1}{30}$.Thus,$v = -30 \ cm$.Differentiating the mirror formula with respect to time $t$: $-\frac{1}{v^2} \frac{dv}{dt} - \frac{1}{u^2} \frac{du}{dt} = 0$.This gives the velocity of the image $v_i = \frac{dv}{dt} = -\left(\frac{v}{u}ight)^2 \frac{du}{dt}$.Given $\frac{du}{dt} = +4 \ cm/s$ (since the object is moving towards the mirror,the distance $u$ is decreasing,but the velocity vector is directed towards the mirror; here we take $u$ as a coordinate,so $u$ changes from $-20$ to $-19$,thus $du/dt = +4 \ cm/s$).$v_i = -\left(\frac{-30}{-20}ight)^2 	imes 4 = -\left(\frac{3}{2}ight)^2 	imes 4 = -\frac{9}{4} 	imes 4 = -9 \ cm/s$.The negative sign indicates that the image is moving in the same direction as the object,which is towards the mirror. However,since the image is real and formed at $30 \ cm$ in front of the mirror,a negative velocity relative to the mirror surface indicates it is moving towards the mirror.

$(i)$ Velocity of image	$(a) \; 2 \; m/s$
$(ii)$ Velocity of image with respect to mirror	$(b) \; 20 \; m/s$
$(iii)$ Velocity of image with respect to object	$(c) \; 11 \; m/s$
$(iv)$ Velocity of image if mirror is stopped	$(d) \; 22 \; m/s$

$A$ luminous point object is moving along the principal axis of a concave mirror of focal length $12 \ cm$ towards it. When its distance from the mirror is $20 \ cm$,its velocity is $4 \ cm/s$. The velocity of the image in $cm/s$ at that instant is:

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