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:
(a)
The maximum possible magnification is obtained when image is formed at near point (d= 25 cm)
Image distance (v) = -d = -25 cm
Focal length (f) = 10 cm
Object distance (u) =?
Using lens formula
1/f = 1/v -1/u
1/u = 1/v -1/f
1/u = 1/-25 - 1/10 = -2-5/50 = -7/50
u = 50/7 = -7.14 cm
Hence lens should be placed 7.14 cm away from them
(b)
Magnification = v/u 25/50/7 = 25 x7/50 = 3.5
(c)
Magnifying power = d/u = 25/50/7 = 25 x 7/50 = 3.5
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