Question

What should be the distance between the object 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

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("A"/"A"_0)`

`= sqrt((6.25)/1)`

= 2.5

But m = `("Image distance (v)")/"Object distnace (u)"`

∴ v = mu = 2.5 u ............(1)

Focal length of the magnifying glass, f = 10 cm

According to the lens formula, we have the relation:

`1/"f" = 1/"v" - 1/"u"`

`1/10 = 1/(2.5"u") - 1/"u" = 1/"u" (1/2.5 - 1/1) = 1/"u" ((1-2.5)/2.5)`

∴ u = `- (1.5  xx 10)/2.5` = −6 cm                   

And v = 2.5 u

= 2.5 × 6 = −15 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.