What should be the distance between the object in Exercise 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?
Given:
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:
…(1)
Where, m = magnification
A = Area of virtual image
A0 = Area of each square
Putting values in equation (1) we get,
m = 
⇒ m = 2.5
Also,
m = Image distance(v)/Object distance(u)
m = v/u
Therefore, v = m × u
= 2.5u …(2)
The magnifying glass has a focal length of, f = 10 cm
Applying the lens formula for lens we have:
…(3)
Where, f0 = focal length of the objective lens
v0 = Distance of image formation
u0 = Distance of object from objective
Putting values in the equation (3) we get,
⇒ 
⇒ u = -(1.5 × 10) / 2.5
Therefore, u = -6 cm
And v = 2.5u
⇒ v = 2.5 × 6
⇒ v = -15 cm
The virtual image cannot be seen by the eyes distinctly, because the image is formed at a distance of 15 cm, which is less than the near point (i.e. 25 cm) of a normal eye.