Question

A 5.0 diopter lens forms a virtual image which is 4 times the object placed perpendicularly on the principal axis of the lens. Find the distance of the object from the lens.

Answer 1

Given,
Power of the lens (P) = $\frac{1}{\mathrm{Focal}\mathrm{length}}$ = 5.0 D
The height of the image is four times the height of the object.
i.e.
$h_i=4h_0,\frac{h_i}{h_0}=4$
We know magnification (m) is also given by
$m=\frac{h_i}{h_0}=\frac{v}{u}\Rightarrow \frac{v}{u}=4,v=4u$
The lens maker formula is given by

$\frac{1}{v}-\frac{1}{u}=\frac{1}{f}$
Here, v is the image distance and u is the object distance.
Now,
$$\begin{gathered} \frac{1}{f}=P=5\mathrm{and}f=\frac{1}{5}\mathrm{m}=20\mathrm{cm} \\ \frac{1}{4u}-\frac{1}{u}=\frac{1}{20} \\ -\frac{3}{4u}=\frac{1}{20} \\ \Rightarrow u=-\frac{60}{4}=-15\mathrm{cm} \end{gathered}$$

Hence, the required object distance is 15 cm.