In a compound microscope the focal length of objective lens is $1.2 c m$ and focal length of eye piece is $3.0 c m$ . When object is kept at $1.25 c m$ in from of objective, final image is formed at infinity. Magnifying power of the compound microscope should be :

400

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200

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150

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100

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Solution

The correct option is A $200$
Given $f_{e} = 3 c m$ and
$f_{o} = 1.2 c m$
$u_{o} = 1.25 c m$
$v_{e} = \infty$

From , $\frac{1}{u_{o}} + \frac{1}{v_{o}} = \frac{1}{f_{o}}$ ,

we get $\frac{1}{v_{o}} = \frac{1}{1.2} - \frac{1}{1.25}$

$v_{o} = 30$ cm

Magnification of microscope $m = \frac{v_{o}}{u_{o}} (\frac{D}{f_{e}})$

$m = \frac{30}{1.25} (\frac{25}{3}) = 200$ .

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In a compound microscope the focal length of objective lens is 1.2 cm and focal length of eye piece is 3.0 cm. When object is kept at 1.25 cm in from of objective, final image is formed at infinity. Magnifying power of the compound microscope should be :

The correct option is A 200Given fe=3cm and fo=1.2cm uo=1.25cm ve=∞From , 1uo+1vo=1fo,we get 1vo=11.2−11.25vo=30 cm Magnification of microscope m=vouo(Dfe)m=301.25(253)=200 .

In a compound microscope the focal length of objective lens is $1.2 c m$ and focal length of eye piece is $3.0 c m$ . When object is kept at $1.25 c m$ in from of objective, final image is formed at infinity. Magnifying power of the compound microscope should be :