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(D) Given: Object distance $u = -20\,m$,Focal length $f = 0.3\,m$,Speed of object $v_o = \frac{du}{dt} = 5\,m/s$ (moving away,so $u$ becomes more nega

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$1.16	imes 10^{-3}\,m/s$ towards the lens

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(D) Given: Object distance $u = -20\,m$,Focal length $f = 0.3\,m$,Speed of object $v_o = \frac{du}{dt} = 5\,m/s$ (moving away,so $u$ becomes more negative,$\frac{du}{dt} = 5\,m/s$).Using the lens formula: $\frac{1}{v} - \frac{1}{u} = \frac{1}{f}$.$\frac{1}{v} - \frac{1}{-20} = \frac{1}{0.3} \Rightarrow \frac{1}{v} = \frac{10}{3} - \frac{1}{20} = \frac{200 - 3}{60} = \frac{197}{60}$.So,$v = \frac{60}{197}\,m$.Differentiating the lens formula with respect to time $t$: $-\frac{1}{v^2} \frac{dv}{dt} + \frac{1}{u^2} \frac{du}{dt} = 0$.$\frac{dv}{dt} = \left( \frac{v}{u} ight)^2 \frac{du}{dt}$.Substituting the values: $\frac{dv}{dt} = \left( \frac{60/197}{-20} ight)^2 	imes 5 = \left( -\frac{3}{197} ight)^2 	imes 5$.$\frac{dv}{dt} = \frac{9}{38809} 	imes 5 \approx 0.00116\,m/s$.Since the sign is positive,the image moves in the direction of light,i.e.,away from the lens.

An object is at a distance of $20\,m$ from a convex lens of focal length $0.3\,m$. The lens forms an image of the object. If the object moves away from the lens at a speed of $5\,m/s$,the speed and direction of the image will be

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