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
