An object 5 cm in length is held 25 cm away

solution:

Given, $\mathrm{h}_{1}=+5 \mathrm{~cm} ; \mathrm{u}=-25 \mathrm{~cm}$

$\mathrm{f}=+10 \mathrm{~cm}$

By lens equation,

$\frac{1}{v}-\frac{1}{u}=\frac{1}{f}$

or $\frac{1}{v}-\frac{1}{(-25)}=\frac{1}{+10}$

or $\frac{1}{v}+\frac{1}{25}=\frac{1}{+10}$

or $\frac{1}{v}=\frac{1}{10}-\frac{1}{25}=\frac{5-2}{50}=\frac{3}{50}$

or $\mathrm{v}=+50 / 3=+16.67 \mathrm{~cm}$

Magnification, $\mathrm{m}=+\frac{\mathrm{v}}{\mathrm{u}}=\frac{+(50 / 3)}{(-25)}=-\frac{2}{3}$

Now, $m=\frac{h_{2}}{h_{1}}$

or $\quad-\frac{2}{3}=\frac{h_{2}}{h_{1}}$

or $h_{2}=-\frac{2}{3} h_{1}=-\frac{2}{3} \times 5=-\frac{10}{3}$

or $\mathrm{h}_{2}=-3.33 \mathrm{~cm}$

The image is real, inverted and diminished.