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Q. An object is placed at 10 cm in front of a concave mirror of radius of curvature 15 cm. The magnification and nature of the image is

Ray Optics and Optical Instruments

Solution:

$f=\frac{R}{2}=\frac{-15}{2}=-7.5 cm$
and $u=-10 cm , \therefore \frac{1}{v}+\frac{1}{(-10)}=\frac{1}{-7.5}$
$\Rightarrow v=-30 cm$
The -ve value of $v,$ means image is real.
$m=-\frac{v}{u}=\frac{-(30)}{-10}=-3$
The image is inverted and magnified 3 times.