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(B) For a lens,the magnification $m = \frac{v}{u} = \frac{f}{f+u}$.The velocity of the image $V_i$ with respect to the lens is given by $V_i = m^2 V_o

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decreasing

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(B) For a lens,the magnification $m = \frac{v}{u} = \frac{f}{f+u}$.The velocity of the image $V_i$ with respect to the lens is given by $V_i = m^2 V_o$,where $V_o$ is the velocity of the object.Since the object is moving towards the lens,$u$ is negative. Let $u = -x$,where $x$ is the distance from the lens.Then $m = \frac{f}{f-x}$.The velocity of the image $V_i = \left(\frac{f}{f-x}ight)^2 V_o$.The relative velocity between the object and the image is $V_{	ext{rel}} = |V_o - V_i| = V_o \left| 1 - \left(\frac{f}{f-x}ight)^2 ight|$.As the object approaches from infinity $(x 	o \infty)$,$m 	o 0$,so $V_i 	o 0$ and $V_{	ext{rel}} 	o V_o$.As the object approaches the lens ($x$ decreases),the term $\left(\frac{f}{f-x}ight)^2$ increases,making $V_i$ increase.Consequently,the relative velocity $V_{	ext{rel}} = V_o(1 - m^2)$ decreases as the object gets closer to the lens.

$A$ real object approaches a fixed diverging lens with a constant velocity from infinity along the principal axis. The relative velocity between the object and its image will be

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