An object approaches a convergent lens from t | vedclass

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

Vedclass · Accepted Answer

Moves away from the lens with non-uniform acceleration.

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(C) For a convergent lens,the lens formula is given by $\frac{1}{v} - \frac{1}{u} = \frac{1}{f}$.Differentiating with respect to time $t$,we get $-\frac{1}{v^2} \frac{dv}{dt} + \frac{1}{u^2} \frac{du}{dt} = 0$,which implies $\frac{dv}{dt} = (\frac{v}{u})^2 \frac{du}{dt}$.Since the magnification $m = \frac{v}{u} = \frac{f}{f+u}$,the velocity of the image is $v_i = m^2 v_o$,where $v_o$ is the velocity of the object.As the object approaches the focus $(u 	o -f)$,the magnification $m$ increases towards infinity.Since $v_i = m^2 v_o$,the image velocity $v_i$ increases rapidly as the object approaches the focus.Because the velocity of the image is changing with time,the image moves 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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