What should be the distance between the object in Exercise 9.30

Solution:

Area of the virtual image of each square, A = 6.25 mm2

Area of each square, A0 = 1 mm2

Hence, the linear magnification of the object can be calculated as:

$m=\sqrt{\frac{A}{A_{0}}}$

$=\sqrt{\frac{6.25}{1}}=2.5$

But $m=\frac{\text { Image distance }(v)}{\text { Object distance }(u)}$

$\therefore v=m u$

$=2.5 u$    ...(1)

Focal length of the magnifying glass, f = 10 cm

According to the lens formula, we have the relation:

$\frac{1}{f}=\frac{1}{v}-\frac{1}{u}$

$\frac{1}{10}=\frac{1}{2.5 u}-\frac{1}{u}=\frac{1}{u}\left(\frac{1}{2.5}-\frac{1}{1}\right)=\frac{1}{u}\left(\frac{1-2.5}{2.5}\right)$

$\therefore u=-\frac{1.5 \times 10}{2.5}=-6 \mathrm{~cm}$

And $v=2.5 u$

$=2.5 \times 6=-15 \mathrm{~cm}$

The virtual image is formed at a distance of 15 cm, which is less than the near point (i.e., 25 cm) of a normal eye. Hence, it cannot be seen by the eyes distinctly.