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(A) From the lens equation,we have $\frac{1}{v} - \frac{1}{u} = \frac{1}{f}$.Differentiating the equation with respect to time $t$,we get $\frac{dv}{d

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amplitude $A/9$,out of phase with the oscillations of the lens

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(A) From the lens equation,we have $\frac{1}{v} - \frac{1}{u} = \frac{1}{f}$.Differentiating the equation with respect to time $t$,we get $\frac{dv}{dt} = \frac{v^2}{u^2} \left( \frac{du}{dt} \right) = m^2 \left( \frac{du}{dt} \right)$,where $m = \frac{v}{u}$ is the magnification.For small oscillations,the amplitude of the image $\Delta v$ is related to the amplitude of the object $\Delta u$ by $\Delta v = m^2 \Delta u$.Given $u = -20 \,cm$ and $f = 5 \,cm$,the magnification $m$ is given by $m = \frac{f}{f+u} = \frac{5}{5-20} = \frac{5}{-15} = -\frac{1}{3}$.Since the lens is oscillating with amplitude $A$ relative to the object,we have $\Delta u = A$. Thus,the amplitude of the image is $\Delta v = m^2 A = \left( -\frac{1}{3} \right)^2 A = \frac{A}{9}$.When the lens moves towards the object,the object distance $u$ decreases (becomes less negative). Since $v = \frac{fu}{f+u}$,as $u$ increases towards $0$,$v$ also changes. Specifically,for a real object placed at $u = -20 \,cm$ $(|u| > f)$,the image is real and formed on the right side. If the lens moves towards the object,the effective distance between the object and the lens decreases,causing the image to move away from the lens. Thus,the oscillations of the image are out of phase with the oscillations of the lens.

$A$ point object is placed $20 \,cm$ to the left of a convex lens of focal length $f=5 \,cm$. The lens is made to oscillate with a small amplitude $A$ along the horizontal axis. The image of the object will also oscillate along the axis with:

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