Question

(a) At what distance should the lens be held from the figure in Exercise 24 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.

Answer 1

(a) Magnifying power of a lens is maximum if the image is at the near point. 

v = – 25 cm, f = 10 cm

As 1/v - 1/u = 1/f 

∴ 1/u = 1/v - 1/f = 1/-25 - 1/10

= (-2 - 5)/50 = - 7/50

∴ u = - 50/7 = - 7.14 cm.

(b) Magnitude of magnification = v/u = 25/7.14 = 3.5

(c) The magnification is equal to the magnifing power is this case because the image is formed at 25 cm.