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(A) For a lens, magnification $m = \frac{v}{u}$.Given $m = \pm 4$ and $f = 20 \,cm$.Case $1$: Real image $(m = -4)$.$m = \frac{v}{u} = -4 \Rightarrow

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$25 \,cm$ when the image is real

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(A) For a lens, magnification $m = \frac{v}{u}$.Given $m = \pm 4$ and $f = 20 \,cm$.Case $1$: Real image $(m = -4)$.$m = \frac{v}{u} = -4 \Rightarrow v = -4u$.Using lens formula $\frac{1}{v} - \frac{1}{u} = \frac{1}{f}$:$\frac{1}{-4u} - \frac{1}{u} = \frac{1}{20} \Rightarrow \frac{-1-4}{4u} = \frac{1}{20} \Rightarrow \frac{-5}{4u} = \frac{1}{20}$.$4u = -100 \Rightarrow u = -25 \,cm$.So, the object is at a distance of $25 \,cm$ when the image is real.Case $2$: Virtual image $(m = +4)$.$m = \frac{v}{u} = 4 \Rightarrow v = 4u$.Using lens formula $\frac{1}{v} - \frac{1}{u} = \frac{1}{f}$:$\frac{1}{4u} - \frac{1}{u} = \frac{1}{20} \Rightarrow \frac{1-4}{4u} = \frac{1}{20} \Rightarrow \frac{-3}{4u} = \frac{1}{20}$.$4u = -60 \Rightarrow u = -15 \,cm$.So, the object is at a distance of $15 \,cm$ when the image is virtual.

An object is placed at a certain distance to the left of a convex lens of focal length $20 \,cm$. Find the distance of the object if the image obtained is magnified by $4$ times.

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