Questions
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?
a
75
b
110
c
140
d
25
detailed solution
Correct option is C
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≈140Talk to our academic expert!
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