Q. 30 A4.8( 4 Votes )

# 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?

Answer :

Given:

Image distance, v = −d = −25 cm

Focal length, f = 10 cm

Object distance = u

When the image is formed at the near point or at the minimum distance of distinct vision, we get the maximum possible magnification.

Applying the lens formula we have:

…(1)

Where, f_{0} = focal length of the objective lens

v_{0} = Distance of image formation

u_{0} = Distance of object from objective

Putting values in the equation (1) we get,

⇒

⇒ u = -50/7

⇒ u = -7.14 cm

Hence, the lens should be kept 7.14 cm away in order to view the squares distinctly.

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PREVIOUSA card sheet divided into squares each of size 1 mm2 is being viewed at a distance of 9 cm through a magnifying glass (a converging lens of focal length 9 cm) held close to the eye.(a) What is the magnification produced by the lens? How much is the area of each square in the virtual image?(b) What is the angular magnification (magnifying power) of the lens?(c) Is the magnification in (a) equal to the magnifying power in (b)? Explain.NEXTWhat is the magnification in this case?

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