(a) At what distance should the lens be held from the figure in Exercise 9.29 in order to view the squares distinctly with the maximum possible magnifying power?
(b) What is the magnification in this case?
(c) Is the magnification equal to the magnifying power in this case? Explain.

Explanation

When image is obtained at near point, then magnifying power becomes maximum.
v = - 25 cm , f = + 10 cm , u = ?
Lens formula ,
`1/f = 1/v - 1/u `
` therefore 1/u = 1/v - 1/f `
`therefore 1/u = 1/(-25) -(1)/(10) = (-2 -5)/(30) = (-7)/(50)`
`therefore u = - (50 )/(7) cm ~~ - 7.14 cm `
`therefore ` Lens should be palced at 7.14 cm for squares.