Question

Use the mirror education to deduce that:
An object placed between f and 2 f of a concave mirror produces a real image beyond 2f.

Answer 1

For a concave mirror, the focal length (f) is negative.
$\therefore F<0$
When the object is placed on the left side of the mirror, the object distance (u) is negative.
$\therefore u<0$
For image distance v, we can write the lens formula as:
$\frac{1}{v}-\frac{1}{u}=\frac{1}{f}$
$\frac{1}{v}=\frac{1}{f}-\frac{1}{u}\dots \dots \left(1\right)$
The object lies between f and 2f.
$\therefore 2f<u<f$ ($\because$ u and are negative)
$\frac{1}{2f}>\frac{1}{u}>\frac{1}{f}$
$-\frac{1}{2f}<-\frac{1}{u}<-\frac{1}{f}$
$\frac{1}{f}-\frac{1}{2f}<\frac{1}{f}-\frac{1}{u}<0\dots \dots \left(2\right)$
Using equation (1), we get:
$\frac{1}{2f}<\frac{1}{v}<0$
$\therefore \frac{1}{v}$ is negative, i.e., v is negative
$\frac{1}{2f}<\frac{1}{v}$
$2f>v$
$-v>-2f$
Therefore, the image lies beyond 2f.