A convex lens $\mathrm{T}$ is used to form an image whose size is one fourth that of size of object. Then…
A convex lens $\mathrm{T}$ is used to form an image whose size is one fourth that of size of object. Then the object distance is
$2\mathrm{f}$
$5 \mathrm{f}$
$4 \mathrm{f}$
$3 \mathrm{f}$
Solution
Magnification, $\mathrm{m}=-\frac{1}{4}$
(Since the image is real and inverted, it is taken as negative)
$\begin{aligned}
& \therefore \mathrm{m}=\frac{\mathrm{v}}{\mathrm{u}}=-\frac{1}{4} \\
& \mathrm{v}=-\frac{\mathrm{u}}{4}
\end{aligned}$
Lens formula, $\frac{1}{\mathrm{v}}-\frac{1}{\mathrm{u}}=\frac{1}{\mathrm{f}}$
$\begin{aligned}
& -\frac{4}{u}-\frac{1}{u}=\frac{1}{f} \\
& u=-5 f
\end{aligned}$
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