Question

A convex lens produces a double size real image when an object is placed at a distance of 18 cm from it. Where should the object be placed to produce a triple size real image?

Answer 1

Given,
Object distance, u = $-$18 cm
Size of the image is two times the size of the object
i.e., h₁ = 2h₀,
We know that,
magnification is given by, m = $\frac{v}{u}$
$\Rightarrow \frac{h_1}{h_0}=\frac{v}{u}=2$
∴ v = 2u = 36 cm.

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

$\frac{1}{36}+\frac{1}{18}=\frac{1}{f}$

$\Rightarrow \frac{1}{f}=\frac{3}{36}$

⇒ f = 12 cm.

For, h₁ = 3h₀

$\frac{h_i}{h_0}=\frac{v}{u}$

$\Rightarrow \frac{v}{u}=3\Rightarrow v=3u$

$-\frac{1}{3u}-\frac{1}{u}=\frac{1}{12}$

u = $-$16 cm.

Therefore, the object should be placed at a distance of 16 cm in front of the lens.