Question

An object of 5.0 cm height is placed at a distance of 20 cm in front of a concave mirror of radius of curvature 30 cm. Find the position of the image, its nature and size.

Answer 1

Radius of curvature (R) = 30 cm
$f=\frac{R}{2}=\frac{30}{2}=15cm$
Focal distance for concave mirror, $f=-15cm$
Object distance, $u=-20cm$
Height of the object, $h_o=5cm$
According to mirror formula,
$\frac{1}{v}+\frac{1}{u}=\frac{1}{f}$
$\frac{1}{v}=\frac{1}{f}-\frac{1}{u}$
$\frac{1}{v}=\frac{1}{-15}+\frac{1}{20}=\frac{-4+3}{60}=\frac{-1}{60}$
Image distance,$v=\frac{-60}{1}=-60cm$
Image is real , inverted and in front of the mirror.
m = $\frac{h_i}{h_0}=-\frac{v}{u}$
$\frac{h_i}{5}=\frac{60}{-20}$
$h_i=-15cm$

Image formed is inverted as size comes out to be negative.
Image is real as it is forming in front of the mirror.
Size of image is 15 cm.