Question

An object of height 6 cm is placed perpendicular to the principal axis of a concave lens of focal length 5 cm. Use lens formula to determine the position, size and nature of the image if the distance of the object from the lens is 10 cm.

Answer 1

We have height of object, $h_1=6cm$, focal length of lens, f = -5 cm and object distance, u = - 10 cm
Using lens formula have $\frac{1}{v}-\frac{1}{u}=\frac{1}{f}$
$\Rightarrow \frac{1}{v}-\frac{1}{(-10)}=\frac{1}{(-5)}\Rightarrow \frac{1}{v}+\frac{1}{10}=-\frac{1}{5}\Rightarrow \frac{1}{v}=-\frac{1}{5}-\frac{1}{10}$
$\Rightarrow v=-\frac{10}{3}=-3.33cm$
Magnification, $M=\frac{v}{u}=-\frac{10}{3}\times (-\frac{1}{10})=\frac{1}{3}$
Again, Magnification, $M=\frac{v}{u}=\frac{h_2}{h_1}\Rightarrow \frac{h_2}{6}=\frac{1}{3}\Rightarrow h_2=\frac{6}{3}=2cm$

Thus the image will be formed in front of the lens at a distance of 3.33 cm from the lens, virtual and erect of size 2 cm.