Question

(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.

Answer 1

(a) for maximum magnifying power, the image should be at least distance of distinct vision. i.e., 25cm.

using lens maker formula,

1/v - 1/u = 1/f

or, 1/-25 - 1/u = 1/+10 [ from exercise 9.29, f = +10cm]

or, -1/u = 1/10 + 1/25 = 7/50

or, u = -50/7 = -7.14 cm

so, lens should be held 7.14 cm from the figure in Exercise 9.29.

(b) linear magnification in the situation of maximum magnifying power,

magnification, m = height of image/height of object = v/u

or, m = (-25)/(-7.14) = 3.5

(c) maximum magnifying power in the same situation.

$m_{max}$ = (1 + D/fe)

here, D = 25cm, fe = 10cm

so, $m_{max}$ = (1 + 25/10) =3.5

so, it can be observed that in the situation when image is least distance of distinct vision the angular magnification ( magnifying power ) and linear magnification have similar value.