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Q. A concave mirror has focal length $f. A$ convergent beam of light is made incident on it. Then the image distance $v$ is

J & K CETJ & K CET 2015Ray Optics and Optical Instruments

Solution:

We know that for a mirror $\frac{1}{v}=-\frac{1}{u}-\frac{1}{t}$
For convergent, $u$ is positive.
For concave mirror, $f$ is negative.
$\Rightarrow \frac{1}{v}=-\frac{1}{u}-\frac{1}{t}$
As, $u >> f$
$\Rightarrow \left[\frac{1}{v} > \frac{1}{f}\right]$
$\Rightarrow v < f$