(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.

Given: The size of each square is 1  mm 2 , the object distance is 9 cmand the focal length of magnifying glass is 10 cm.

(a)

The lens formula is given as,

1 f = 1 v − 1 u

Where, object distance is u, focal length of converging lens is fand the image distance is v

The maximum possible magnification is obtained when the image is formed at the near point of eye.

So,

v=−25 cm

By substituting the given values in the above expression, we get

1 u = 1 v − 1 f 1 u = 1 −25 − 1 10 = −2−5 50 u=−7.14 cm

Thus, to view the squares distinctly, the lens should be kept 7.14 cmaway from them.

(b)

Magnification is given as,

m=| v u |

By substituting the given values in the above expression, we get

m=| −25 50 7 | = 25×7 50 =3.5

Thus, in this case the magnification is 3.5 times.

(c)

Magnifying power of the lens is given as,

m= d | u |

Where, d is the least distance for distinct vision.

By substituting the given values in the above expression,

m= 25 | 50 7 | = 25×7 50 =3.5

Thus, the magnifying power of the lens is 3.5.

Since, in this case the image is formed at the near point of the eye at 25 cm, the magnifying power is equal to the magnitude of magnification.