The size of the real image produced by a convex lens of focal length $\mathrm{F}$ is ' $\mathrm{m}$ ' times…
The size of the real image produced by a convex lens of focal length $\mathrm{F}$ is ' $\mathrm{m}$ ' times the size of the object. The image distance from the lens is
$\frac{F}{(m-1)}$
$\frac{(m-1)}{F}$
$\mathrm{F}(\mathrm{m}+1)$
$\mathrm{F}(\mathrm{m}-1)$
Solution
From lens formula, $\frac{1}{\mathrm{v}}-\frac{1}{\mathrm{u}}=\frac{1}{\mathrm{~F}}$
For a convex lens, $\mathrm{u}$ is negative, $\mathrm{v}$ and $\mathrm{F}$ are positive.
$\begin{aligned}
& \therefore \quad \frac{1}{\mathrm{v}}-\frac{1}{-\mathrm{u}}=\frac{1}{\mathrm{~F}} \\
& \therefore \quad \frac{1}{\mathrm{v}}=\frac{1}{\mathrm{~F}}-\frac{1}{\mathrm{u}}
\end{aligned}$
Multiplying by $\mathrm{v}$,
$\begin{aligned}
1 & =\frac{\mathrm{v}}{\mathrm{F}}-\frac{\mathrm{v}}{\mathrm{u}} \\
\therefore \quad 1 & =\frac{\mathrm{v}}{\mathrm{F}}-\mathrm{m} \quad \ldots\left(\because \frac{\mathrm{v}}{\mathrm{u}}=\mathrm{m}\right) \\
\therefore \quad \frac{\mathrm{v}}{\mathrm{F}} & =1+\mathrm{m} \\
\therefore \quad \mathrm{v} & =\mathrm{F}(1+\mathrm{m})
\end{aligned}$
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