The focal length of objective and eye piece of microscope is 2 cm and 5 cm . And the distance between the lenses is 20 cm . The distance of object from the objective , when the focal image seen by eye is 25 cm from eye piece . So what will be the magnifying power ?

The focal length of objective and eye piece of microscope is 2 cm and 5 cm . And the distance between the lenses is 20 cm .  The distance of object from the objective , when the focal image seen by eye is 25 cm from eye piece . So what will be the magnifying power ?

Choose single option from these following : -

A ) 32 ,                                             B ) 41 ,

C ) 68 ,                                             D ) None of these ,

Correct Answer : -   B ) 41 ,

Solution : -

Image of eye piece : -  Ve = - 25 cm , 

And  fe = 5 cm .    ∴  fe = focal length of eye piece . 

So , formula for lens is : -

( 1 / fe )  =  ( 1 / Ve )  -  ( 1 / Ue )  .

So  ,   ( 1 / Ue )  =  ( 1 / Ve )  -  ( 1 / fe )  ,

( 1 / Ue )  =  ( 1 / -25 )  -  ( 1 / 5 )  ,

So ,  Ue =  - 25 / 6 .

Vo  =  20 – ( 25 / 6 )  =  95 / 6  cm   ,   And  fo = 2 cm .

So ,

( 1 / Uo )  =  ( 1 / Vo )  -  ( 1 / fo )  ,

( 1 / Uo )  =  ( 1 / -25 )  -  ( 1 / 5 )  ,

Uo  =  - 2.3 cm .  So , it means that object is away from final image .

So , magnifying power : -

M  =  ( Vo / Uo ) ( 1 + d / fe )  ,             ∴  d = 25 .

So simply put the values , we will have answer : -

M = 41  ,  Answer .     

OR