An object approaches a convergent lens from t | vedclass

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(C) For a convex lens,the lens formula is given by $\frac{1}{v} - \frac{1}{u} = \frac{1}{f}$.Differentiating with respect to time $t$,we get $-\frac{1

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Moves away from the lens with non-uniform acceleration.

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(C) For a convex lens,the lens formula is given by $\frac{1}{v} - \frac{1}{u} = \frac{1}{f}$.Differentiating with respect to time $t$,we get $-\frac{1}{v^2} \frac{dv}{dt} + \frac{1}{u^2} \frac{du}{dt} = 0$.This implies the velocity of the image $v_i = \frac{dv}{dt} = (\frac{v}{u})^2 \frac{du}{dt}$.Since $v = \frac{uf}{u+f}$,we have $v_i = (\frac{f}{u+f})^2 v_o$,where $v_o$ is the speed of the object.As the object approaches the focus $(u 	o -f)$,the denominator $(u+f) 	o 0$,which causes the image velocity $v_i$ to increase rapidly towards infinity.Since the velocity is changing with respect to time and position,the acceleration is non-uniform.Therefore,the image moves away from the lens with non-uniform acceleration.

An object approaches a convergent lens from the left of the lens with a uniform speed $5 \ m/s$ and stops at the focus. The image :-

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