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 mm^{2}. Would you be able to see the squares distinctly with your eyes very close to the magnifier?

#### Solution

Area of the virtual image of each square, A = 6.25 mm^{2}

Area of each square, A_{0} = 1 mm^{2}

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.