An object 5.0 cm in length is placed at a distance

solution:

Given, object distance, $\mathrm{u}=-20 \mathrm{~cm}$;

radius of curvature, $\mathrm{R}=+30 \mathrm{~cm}$;

height of object, $h_{1}=+5 \mathrm{~cm}$;

image distance, $\mathrm{v}=? ;$ magnification, $\mathrm{m}=?$

height of image, $\mathrm{h}_{2}=?$

Focal length, $\mathrm{f}=\mathrm{R} / 2=(+30) / 2=+15 \mathrm{~cm}$

Mirror formula,

$\frac{1}{v}+\frac{1}{u}=\frac{1}{f} \quad$ or $\quad \frac{1}{v}+\frac{1}{(-20)}=\frac{1}{(+15)}$

or $\frac{1}{v}=\frac{1}{20}+\frac{1}{15}=\frac{3+4}{60}=\frac{7}{60}$

or $v=+60 / 7 \mathrm{~cm}=+8.57 \mathrm{~cm}$

Now, magnification, $m=-\frac{v}{u}=-\frac{(+60 / 7)}{(-20)}=+\frac{3}{7}$

or $\mathrm{m}=+0.428$

Also, $m=\frac{h_{2}}{h_{1}} \quad$ or $\quad+\frac{3}{7}=\frac{h_{2}}{(+5)}$

or $h_{2}=+\frac{15}{7}=+2.14 \mathrm{~cm}$

The image is located at a distance of 8.57 cm behind the mirror ; it is virtual, erect and diminished of size 2.14 cm.