An object 5 cm in length is held 25 cm away from a converging lens of focal length 10 cm. Draw the ray diagram and find the position, size and the nature of the image formed.
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$f=+10 cm, u =-25 cm$ and $h _0=5 cm$
$\frac{1}{\text{f}}=\frac{1}{\text{v}}-\frac{1}{\text{u}}$
$\frac{1}{\text{v}}=\frac{1}{10}-\frac{1}{15}$
$\frac{1}{\text{v}}=\frac{3}{50}$
$\text{V}=\frac{50}{3} \ \text{cm}$.
The image is real and inverted at a distance of 16.7 cm from the lens on opposite side.
Magnification $(\text{m})=\frac{\text{h}_\text{i}}{\text{h}_0}=\frac{\text{V}}{\text{u}}\frac{\text{h}_\text{i}}{5}$
$=\frac{16.7}{-25}\text{h}_\text{i}$
$=\frac{-10}{3}\text{cm}.$ image is inverted and diminished.
$\frac{1}{\text{f}}=\frac{1}{\text{v}}-\frac{1}{\text{u}}$
$\frac{1}{\text{v}}=\frac{1}{10}-\frac{1}{15}$
$\frac{1}{\text{v}}=\frac{3}{50}$
$\text{V}=\frac{50}{3} \ \text{cm}$.
The image is real and inverted at a distance of 16.7 cm from the lens on opposite side.
Magnification $(\text{m})=\frac{\text{h}_\text{i}}{\text{h}_0}=\frac{\text{V}}{\text{u}}\frac{\text{h}_\text{i}}{5}$
$=\frac{16.7}{-25}\text{h}_\text{i}$
$=\frac{-10}{3}\text{cm}.$ image is inverted and diminished.
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