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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:

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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=1vo−1uo

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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