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(D) Given: Focal length $f = 0.05 \ m$,object distance $u = -0.2 \ m$.Using the lens formula: $\frac{1}{v} - \frac{1}{u} = \frac{1}{f}$.Substituting t

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$\frac{A}{9} 	imes 10^{-2} \ m$

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(D) Given: Focal length $f = 0.05 \ m$,object distance $u = -0.2 \ m$.Using the lens formula: $\frac{1}{v} - \frac{1}{u} = \frac{1}{f}$.Substituting the values: $\frac{1}{v} = \frac{1}{f} + \frac{1}{u} = \frac{1}{0.05} - \frac{1}{0.2} = 20 - 5 = 15 \ m^{-1}$.Thus,$v = \frac{1}{15} \ m$.Differentiating the lens formula $\frac{1}{v} - \frac{1}{u} = \frac{1}{f}$ with respect to time,we get: $-\frac{dv}{v^2} + \frac{du}{u^2} = 0$.This implies $dv = \left( \frac{v^2}{u^2} ight) du$.The amplitude of the image oscillation $A_{image} = |dv|$ is given by $A_{image} = \left( \frac{v}{u} ight)^2 A_{object}$.Given $A_{object} = A \ cm = A 	imes 10^{-2} \ m$.$A_{image} = \left( \frac{1/15}{0.2} ight)^2 	imes A 	imes 10^{-2} \ m = \left( \frac{1}{15 	imes 0.2} ight)^2 	imes A 	imes 10^{-2} \ m = \left( \frac{1}{3} ight)^2 	imes A 	imes 10^{-2} \ m = \frac{A}{9} 	imes 10^{-2} \ m$.

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