An object is moving towards a converging lens | vedclass

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Question

(D) For a converging 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

$u < f$

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(D) For a converging 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 $\frac{dv}{dt} = \left( \frac{v}{u} \right)^2 \frac{du}{dt}$.Let $v_o = \frac{du}{dt}$ be the velocity of the object and $v_i = \frac{dv}{dt}$ be the velocity of the image.Since the object moves towards the lens,$\frac{du}{dt}$ is negative. For the image to also move towards the lens,$\frac{dv}{dt}$ must be negative.However,the term $\left( \frac{v}{u} \right)^2$ is always positive. This implies that for a real image $(v > 0)$,the image moves in the opposite direction to the object.If the object is placed between the focus and the optical center $(u As the object moves from $F$ towards $O$ (i.e.,$u$ decreases from $f$ to $0$),the virtual image moves from $-\infty$ towards $O$. Thus,both the object and the image move towards the lens.Therefore,the condition is $u < f$.

An object is moving towards a converging lens on its axis. The image is also found to be moving towards the lens. Then,the object distance $u$ must satisfy:

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