A convex lens of focal length ' $f$ ' $m$ forms a real, inverted image twice in size of the object. The object distance from the lens in metre is

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Aicharya · Accepted Answer

$\begin{aligned} & 	ext { Focal length }=\mathrm{f} \ & \mathrm{~m}=\frac{\mathrm{v}}{\mathrm{u}}=\frac{\mathrm{h}^{\prime}}{\mathrm{h}}=-2 ...(given)\ 	herefore \quad & \mathrm{v}=-2 \mathrm{u} \end{aligned}$ From lens formula, $\begin{aligned} & \frac{1}{v}-\frac{1}{u}=\frac{1}{f} \ & \frac{1}{-2 u}-\frac{1}{u}=\frac{1}{f} \ & \frac{u+2 u}{2 u^2}=\frac{1}{f} \ & \frac{3 u}{2 u^2}=\frac{1}{f} \ & \frac{1.5}{u}=\frac{1}{f} \ & 	herefore \quad u=1.5 \mathrm{f}\end{aligned}$

A convex lens of focal length ' $f$ ' $m$ forms a real, inverted image twice in size of the object. The…

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