Q: A graph sheet divided into squares each of size 1 mm2 is kept at a distance of 7 cm from a magnifying glass of focal length of 8 cm. The graph sheet is viewed through the magnifying lens keeping the eye close to the lens. Find
(i) the magnification produced by the lens,
(ii) the area of each square in the image formed
(iii) the magnifying power of the magnifying lens. Why is the magnification found in (i) different from the magnifying power?
Sol: (i) u = -7 cm; f = +8 cm; v = ?
For a lens, $\large \frac{1}{v} – \frac{1}{u} = \frac{1}{f} $
$\large \frac{1}{v} – \frac{1}{-7} = \frac{1}{8} $
v = -56 cm
Magnification, M = v/u = (-56)/(-7) = +8
(ii) Each square is of size 1 mm2 ie. its length and breadth are each to 1 mm The virtual image formed has linear magnification 8. So its length and breadth are each equal to 8 mm, The area of the image of
each square = 8 × 8 mm2 = 64 mm2
(iii) Magnifying power of the magnifying glass ie. simple microscope.
$\large m = 1 + \frac{D}{f} $
$\large m = 1 + \frac{25}{8} $
= 4.125
The magnification found in (i) is different from the magnifying power because the image distance in (i) is different from the least distance of distinct vision D.