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 = `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 = 3h_{0
\[\frac{h_i}{h_0} = \frac{v}{u}\]
\[\Rightarrow   \frac{v}{u} = 3             \Rightarrow v = 3u\]}

\[- \frac{1}{3u} - \frac{1}{u} = \frac{1}{12}  \] 
u = - 16 cm. 
Therefore, the object should be placed at a distance of 16 cm in front of the lens.