An object approaches a convergent lens from t | vedclass

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Question

(C) From the lens formula,$\frac{1}{f} = \frac{1}{v} - \frac{1}{u}$.Differentiating with respect to time $t$,we get $0 = -\frac{1}{v^2} \frac{dv}{dt}

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moves away from the lens with a non-uniform acceleration

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(C) From the lens formula,$\frac{1}{f} = \frac{1}{v} - \frac{1}{u}$.Differentiating with respect to time $t$,we get $0 = -\frac{1}{v^2} \frac{dv}{dt} + \frac{1}{u^2} \frac{du}{dt}$.Rearranging,we find the velocity of the image $v_i = \frac{dv}{dt} = \left(\frac{v^2}{u^2}ight) \frac{du}{dt} = m^2 v_o$,where $m = \frac{v}{u}$ is the magnification and $v_o$ is the object speed.Since $m = \frac{f}{f+u}$,the magnification $m$ changes as the object moves towards the focus $(u 	o -f)$.Differentiating the velocity $v_i$ with respect to time to find acceleration $a_i = \frac{dv_i}{dt} = \frac{d}{dt}(m^2 v_o) = 2m v_o \frac{dm}{dt}$.As the object approaches the focus,$m$ changes non-linearly,making $\frac{dm}{dt}$ variable. Thus,the image moves away from the lens with a 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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