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Q.
A converging lens has a focal length of $1\, m$. The minimum distance between a real object and its real image formed by this lens is
Ray Optics and Optical Instruments
Solution:
$\frac{1}{v}-\frac{1}{u}=\frac{1}{f}$
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 $f=1 \Rightarrow \frac{1}{y}+\frac{1}{x}=\frac{1}{f}$
$\Rightarrow \frac{1}{(d-x)}+\frac{1}{x}=1$
$\frac{d}{x d-x^{2}}=1$
$d=x d-x^{2} \text { or } x^{2}-d x +d=0$
For real roots, $d^{2}-4 d \geq 0$
$d(d-4) \geq 0 ; d \geq 4$