Question

(a) At what distance should the lens be held from the card sheet in Exercise 9.22 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) The maximum possible magnification is obtained when the image is formed at the near point (d = 25 )
Image distance, v = -d = -25cm
Focal length, f = 10cm
Object distance = u
According to the lens formula, we have :\(\frac{1}{f}=\frac{1}{v}-\frac{1}{u}\)
\(\frac{1}{u}=\frac{1}{v}-\frac{1}{f}\)

= \(\frac{1}{-25}-\frac{1}{10}\)

= -\(\frac{2-5}{50}\)

=\(-\frac{7}{50}\)

∴ u = \(-\frac{50}{7}\) = - 7.14cm
Hence, to view the squares distinctly, the lens should be kept 7.14cm away from them.
(b) Magnification = |\(\frac{v}{u}\)| = \(\frac{25}{\frac{50}{7}}\) = 3.5
(c) Magnifying power = \(\frac{d}{u}\)=\(\frac{25}{\frac{50}{7}}\) = 3.5 Since the image is formed at the near point ( 25cm ), the magnifying power is equal to the magnitude of magnification.