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# A compound microscope consists of an objective lens of focal length 2.0cm and an eyepiece of focal length 6.25cm separated by a distance of 15cm. How far from the objective should an object be placed in order to obtain the final image at (a) the least distance of distinct vision (25cm), and (b) at infinity? What is the magnifying power of the microscope in each case?

Vikas TU
14149 Points
one year ago
Focal length of objective lens =fo=2 cm
Focal length of eye piece =fe=6.25cm
Distance between both the lenses =L=15cm
Distance of final image from the eye piece=fe= - 25m
Formula to be used : eye piece
1/f=1/v-1/u
1/ve-1/ue=1/fe
1/ue=1ve-1/fe
=1/-25 - 1/6.25
=-1-4/25
=-5/25 = -1/5
As the distance between objective and eye lens =ve+ue=15cm
Distance of the image formed by object lens:
V0=L- | ue| =15-5=10cm
Using lens formula for objective lens:
1/vo-1/uo=1/f0
1/uo=1/fo-1/vo
=1/10-1/2
=1-5/10= -4/10
u0 = --2.5cm
so the object should be 2.5cm in front of convex lens
Magnifying power of compound microscope :
m=-vo/|uo| (1+d/fe)
= -10/2.5[1+ 25/6.25]
=20 [ since d=25cm]
∴Magnifying power of Microscope is 20
b) the final image is formed at INFINITY only if the image formed by the objective is in the focal plane of eye piece.
Thus,
ve= - ∞
ue=fe=6.25cm
Image distance of objective lens :
vo=L-fe
=15-6.25
=8.75cm
Using lens formula:
1/vo-1/uo=1/fo
1/uo=1vo-1/fo
=1/8.75-1/2
=2-8.75/17.5
uo= -17.5/6.75
= -2.59cm
Magnifying power of microscope  is :
m=v0/|uo|  [1+d/fe]
=8.75/2.59 (1+25/6.25]
=13.51
∴Magnifying power of Microscope is 13.51