A compound microscope consists of an objective lens of focal length 2.0cm and an eyepiece of focal length 6.25cm separated by a distance of 15cm .

A compound microscope consists of an objective lens of focal length 2.0cm and an eyepiece of focal length 6.25cm separated by a distance of 15cm . How far from the objective should an object be placed in order to obtain the final image at: (a) the least distance of distinct vision (25cm) , and (b) at infinity? What is the magnifying power of the microscope in each case?

 

Focal length of the objective lens fo=2cm

Focal length of the eyepiece lens fe=6.25cm

Distance between the objective lens and the eyepiece, L=15cm

Least distance of distinct vision,D=25cm

Image distance for the eyepiece, ve=-25cm

Object distance for the eyepiece, is ue

Applying lens formula for eyepiece lens,

1ve-1ue=1fe

1-25-1ue=16.25

ue=-5cm

Image distance from objective lens is v0=L-ue=15-5=10cm

Now applying lens formula for objective lens,

1vo-1uo=1fo

110-1uo=12

uo=-2.5cm

Magnifying power of a compound lens for this case is,

m=vouoDfe+1 

= 102.5256.25+1

= 20

Image distance for the eyepiece, ve=-

Object distance for the eyepiece, is ue

Applying lens formula for eyepiece lens,

1ve-1ue=1fe

1--1ue=16.25

ue=-6.25cm

Image distance from objective lens is v0=L-ue=15-6.25=8.75cm

Now applying lens formula for objective lens,

1vo-1uo=1fo

18.75-1uo=12

uo=-2.59cm

Magnifying power of a compound lens for this case is,

m=vouoDfe 

= 8.752.59256.25

= 13.51

When Final Image is at least distance of distinct vision,

Distance of object from the objective lens isu0=2.5cm

Magnifying power in this case is m=20

When Final Image is at infinity,

Distance of object from the objective lens isu0=2.59cm

Magnifying power in this case is m=13.51

 

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