Q.An object is placed at a certain distance from a convex lens of focal length 20 cm. Find the distance of the object if the image obtained is magnified 4 times.

Step 1. For a lens of focal length $f=20$ cm, magnification $m=v/u$; taking the virtual-image case (object between the lens and its focus, magnifying-glass configuration), $m=+4$, so $v=4u$.

Step 2. Substituting into the lens equation $\dfrac{1}{v}-\dfrac{1}{u}=\dfrac{1}{f}$: $\dfrac{1}{4u}-\dfrac{1}{u}=\dfrac{1}{20}$.

Step 3. Simplifying the left side: $\dfrac{1-4}{4u}=\dfrac{-3}{4u}=\dfrac{1}{20}$, so $u=\dfrac{-3\times20}{4}=-15$ cm.

Step 4. Checking: with $u=-15$ cm, $\dfrac{1}{v}=\dfrac{1}{20}+\dfrac{1}{-15}=\dfrac{3-4}{60}=-\dfrac{1}{60}$, so $v=-60$ cm, and $m=v/u=(-60)/(-15)=4$ (positive, confirming a virtual, erect, magnified image), consistent with the given book answer.

Set m = v/u = +4 (virtual-image case) and substitute into the lens equation to solve for u.

• Using the real-image case (m = -4) instead of the virtual, magnifying-glass case (m = +4) the book's answer corresponds to.
• Sign errors when substituting v = 4u into the lens equation.