Question

A converging lens has a focal length of $1m$. The minimum distance between a real object and its real image formed by this lens is

• A. 0.5 m
• B. 1 m
• C. 2 m
• D. 4 m

Answer 1

Answer: (D)

$\frac{1}{v}-\frac{1}{u}=\frac{1}{f}$
If $x$ and $y$ be the distance of real object and real image respectively, then, $u=-x$ and $v=+y$
Let $x+y=d,$ here $f=1\Rightarrow \frac{1}{y}+\frac{1}{x}=\frac{1}{f}$
$\Rightarrow \frac{1}{(d-x)}+\frac{1}{x}=1$
$\frac{d}{xd-x^2}=1$
$d=xd-x^2\text{ or }{x}^2-dx+d=0$
For real roots, $d^2-4d\ge 0$
$d(d-4)\ge 0;d\ge 4$