Question

A convex lens of focal length 0.10 m is used to form a magnified image of an object of height 5 mm placed at a distance of 0.08 m from the lens. Calculate the position, nature and size of the image.

Answer 1

Here, $u=-0.08m=-8cm$

$f=0.1m=10cm$

and $h_o=5mm$

Using lens formula, $\frac{1}{v}-\frac{1}{u}=\frac{1}{f}$

$\Rightarrow \frac{1}{v}-\frac{1}{-8}=\frac{1}{10}$

$\Rightarrow \frac{1}{v}=\frac{1}{10}-\frac{1}{8}=\frac{-1}{40}$

$\Rightarrow v=-40cm$

Hence, the image is formed at a distance of 40 cm on the same side as the object.

Now, $m=\frac{v}{u}=\frac{-40}{-8}=5$

$\Rightarrow \frac{h_i}{5mm}=5$

$\Rightarrow h_i=25mm$

Hence, the image is virtual, erect and magnified.