Define the magnifying power of compound microscope.

Explanation

Compound microscope : It is a device used to see the tiny particles/objects which can not be seen with naked eye.
Principle : When an object is placed between Fand 2F in front of a convex lens (o) then the real, inverted and magnified image is formed on the other side of the lens. This image acts as an object for eye piece. Position of eye piece is so adjusted that object image formed by o) lies within the focal length of eye piece of large aperture. The final image produced by eye piece is virtual, erect and highly magnified.
Construction : it consists of two convex lenses called objective and eye piece.
(i) Objective lens : A lens lying towards the object is called objective lens.
Objective lens has small focal length and small aperture.
(ii) Eye piece : The lens, which is towards the eye i.e., through whiclı observer observes the object is known as eye piece.
Its focal length and aperture are large as compared to the objective lens.
These lenses are mounted at the two ends of two tubes. With the help of rack pinion arrangement, distance of object from object lens is varied.

Magnifying power or Angular magnification : It is defined as the ratio of the angle subtended by the final image at the eye of the angle subtended by the object seen directly when both are placed at least distance of distinct vision.
Let `anglePC_(2)A..=alpha=` Angle subtended by the object at eye when placed at least distance of distant vision
and `angleA..C_(2)B.=beta=` Angle subtended by the final image at the eye when placed at the distance of distinct vision.
`:. M.P. =(beta)/(alpha)" " ...(i)`
Since `alpha` and `beta` are small angles,
then `M.P.=(tan beta)/(tan alpha)" "...(ii)`
From `DeltaA..C_(2)B., tan beta=(A.B.)/(C_(2)A.)` and from `DeltaPC_(2)A.., tan alpha=(PA.)/(C_(2)A.)`
Hence eqn. (ii) becomes
`M.P.=(A.B.//C_(2)A.)/(PA.//C_(2)A.)=(A.B.)/(PA.)" " ...(iii)`
Since `PA..=AB " " :.M.P.=(A.B.)/(AB) " " ...(iv)`
Multiplying and dividing R.H.S. of eqn. (iv) by A.B, we get
`M.P.=(A.B.)/(AB) xx (AB)/(AB)=(A.B.)/(AB) xx (AB)/(AB) " " ...(v)`
But `(AB)/(AB) =("Size of image")/("Size of object")` (for objective lens) `=m_(o)`
and magnification of eye lens `=m_(e )=(A.B.)/(AB)`
Put values of `m_(o)` and `m_(e )` in eqn. (v)
`M.P.=m_(o) xx m_(e )" "...(vi)`
Now `m_(o)=(A.B.)/(AB)=(v_(0))/(-u_(0)) " "...(vii)`
where `v_(o)=` distance of image A.C from objective lens.
`u_(0)=` distance of object AB from objective lens
Also `m_(e )=(A.B.)/(AB)=(v)/(u_(e ))`
where v = Distance of final image A" B. from eyepiece
But v = D because final image is formed at least distance of distinct vision.
So eq. (viii) becomes
`m_(e )=(D)/(u_(e ))`
Using lens formula for eye piece,
`-(1)/(u)+(1)/(v)=(1)/(f_(e ))`
Applying new cartesian sign convention `u= -u_(e ), v= -D`
`:. -(1)/((-u_(e )))+(1)/((-D))=(1)/(f_(e ))`
Multiplying both sides by D, we get
`(D)/(u_(e )) -1=(D)/(f_(e ))`
`rArr (D)/(u_(e ))=1+(D)/(f_(e ))`
`rArr m_(e )=(D)/(u_(e )) = 1+(D)/(f_(e ))`
Put values of `m_(o)` and `m_(e )` in eqn. (vi)
`M.P.= (v_(0))/(-u_(0))(1+(D)/(f_(e )))`
If the object AB is placed very close to the focus of the objective lens i.e., `u_(o)=f_(o)` and image A.B. is formed very close to the eye lens i.e.,
`v_(o)=f_(o)=L=` Distance between objective and eye piece = Length of microscope
`rArr M.P= -(L)/(f_(o)) (1+(D)/(f_(e )))`
which is expression for magnifying power of compound microscope.
Magnifying power of a compound microscope can be increased by decreasing the focal length of both the lenses.