Question

What should be the distance between the object in Exercise 25 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?

Answer 1

Magnification = √(Area of the virtual image, (square)/Area of the objec = t (square))

m = √(6.25/1) = 2.5

As, m = v/u ∴ v = mu = + 2.5u

As, 1/v - 1/u = 1/f ⇒ 1/2.5u - 1/u = 1/10

⇒ (1 - 2.5)/2.5u = 1/10 ⇒ - 1.5/2.5u = 1/10

∴ 2.5u = - 1.5 x 10 ⇒ u = - 15/2.5 = - 6cm.

v = 2.5u = (2.5) (– 6) = –15 cm

As the virtual image formed is closer than normal near point (25 cm), therefore it cannot be seen distinctly.