An angular magnification (magnifying power) of `30 X` is desired using an objective of focal length `1.25 cm` and an eye piece of focal length `5 cm`. How will you set up the compound microscope ?

Explanation

Focal length of hte objective lens, `f_(0)`= 1.25 cm
Focal length of the eyepiece, fe=5 cm
long distance of distinct vision, d=25cm
Angular magnification of the compound microscope = 30X
Total magnifying power of the compound microsope, m=30
The angular magnification of the eyepiece is given by the relation,
`m_(e) = (1+d/f_(e))`
=`(1+25/5)=6`
the angular magnification of the objective lens (mo) is related to me as:
`m_(e)m_(0) = m`
`m_(0)= m/m_(e)`
`=30/6=5`
We also have the relation:
`m_(0) = ("Image distance for the objective lens" (v_(0)))/("object distance for the objective lens"(-u_(0)))`
`5=v_(0)/-u_(0)`............(i)
Applying the lens formula for the objective lens:
`1/(f_(0)) = 1/v_(0)-1/u_(0)`
`1/1.25= 1/-5u_(0)-1/u_(0) = -6/(5u_(0)`
`therefore u_(0) = -6/5 xx 1.25 = -1.5cm`
And `v_(0) = -5u_(0)`
`-5 xx (-1.5)=7.5cm`
The object should be placed 1.5 cm away from the objective lens to obtain the desired magnification.
Applying the lens formula for the eyepiece.
`1/v_(c)-1/u_(c) =1/f_(c)`
Where,
`v_(c)` = Image distance for the eyepiece= `-d = -25cm`
`u_(c)` = object distance for the eyepiece.
`1/u_(c) =1/v_(c) -1/f_(c)`
`=-1/25 -1/5=-6/25`
`therefore u_(c) =-4.17 cm`
Separation between the objective lens and the eyepiece `=|u_(e)|+|v_(0)|`
=`4.17+7.5`
=`11.67cm`
Therefore, the separation between the objective lens and the eyepiece should be 11.67cm