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Q.

A microscope has an objective of focal length 1.5 crn and eye piece of focal length 2.5 cm. If the distance between objective and eyepiece is 25 cm. what is the approximate value of magnification produced for relaxed eye?

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a

75

b

110

c

140

d

25

answer is C.

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Detailed Solution

Length  of  the  tube  is  L=vo+fevo=L−fe=25−2.5=22.5  cmNow  applying  1vo−1uo=1fowe  have  122.5−1uo=11.5∴uo≈1.6 cm∴M=vouo×Dfe=22.51.6252.5≈140

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