What should be the distance between the object (in Question 9.30) and the magnifying glass if the virtual image of each square in the figure is to have an area of 6.25 mm². Would you be able to see the squares distinctly with your eyes very close to the magnifier?

Solution

Here,
Magnification in area = 6.25
∴ Linear magnification,

m = \frac{v}{u} v = mu = 2.5 u

Using formula,

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

∴

\frac{1}{2.5 u} - \frac{1}{u} = \frac{1}{10}

\frac{1 - 2.5}{2.5 u} = \frac{1}{10}

2.5 u = - 15

i.e.,

u = - 6 cm

∴

v = 2.5 u = 2.5 (- 6) = - 15 cm

The virtual image is at 15 cm; whereas distance of distinct vision is 25 cm, therefore, the image cannot be seen distinctly by the eye.

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