Question

At what distance from a convex lens of focal length $0.15m$ must an object be placed so that the image must be twice the size of the object, when the image is
a) real
b) virtual

Answer 1

The magnification is given as

$m=\frac{h_i}{h_0}$

As it is given that the height of the image is twice that of the object

$m=2$

We know that

$\frac{h_i}{h_0}=-\frac{u_i}{v_0}=2$

By using the formula

$\frac{1}{f}=\frac{1}{u_0}+\frac{1}{v_i}$

$\frac{1}{15}=\frac{1}{u_0}-\frac{1}{2u_0}$

Hence for virtual image the distance of the object is $7.5cm$ and the distance of the image should be $-15cm$

For real image

By using the formula

$m=-\left(\frac{-u_i}{v_0}\right)$

Hence the distance of the object is $7.5cm$ and the distance of the image should be $-15cm$