What should be the distance between the object and 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 ?

Explanation

Area of the virtual image of each square, `A=6.25 mm^(2)`
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=("Imaage distance"(v))/("Object distance"(u))`
`therefore v=mu`
`=2.5u....................(i)`
Focal length of the magnifying glass, f=10cm
According to the lens formula, we have the relations,
`1/f=1/v-1/u`
`1/10= 1/2.5u -1/u=1/u(1/2.5-1/1)=1/u((1-2.5)/(2.5))=-6cm`
And v=2.5 u
`=2.5 xx 6=-15cm`
The virtual image is formed at a distance of 15cm, which is less than the near point (i.e=25 cm) of a normal eye. Hence, it cannot be seen by the eyes distinctly.