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Question

The focal length of the objective and eye piece of a compound microscope are 1 cm and 5 cm respectively. An object is placed at a distance of 1.1 cm from objective. If final image is formed by at least distance vision, the magnifying power is:

A
60
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B
50
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C
40
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D
None of these
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Solution

The correct option is B 50
Given that,
Focal length of objective fo=1cm
Focal length of eyepiece fe=5cm
Distance of the object uo=1.1cm
Distinct vision = D

Now,
The magnifying power of a compound microscope is the product of the linear magnification of the objective and the magnifying power of the eyepiece.
M.P=Mo×Me
M.P=(v0ve)×(Du0).....(I)

Now, for the objective lens
1fo=1vo1uo
vo=uofouo+fo
vouo=fouo+fo

Now, put the value in equation (I)
M.P=(fouo+fo)×(Duo)

Now, because uo>fo
So,
M.P=(fouo+fo)×(Duo)

Now, the distance of distinct vision, D may be taken as 25 cm.
So,
M.P=(fouo+fo)×(Duo)
M.P=(11.1+1)×255
M.P=50

The negative sign indicates that the image is inverted
Hence, the magnifying power is 50


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