. 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?

 

Best Answer

Explanation:

  • 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

objective lens

  • Image distance for the eyepiece, ve=-25cm
  • Object distance for the eyepiece, is ue
  • Applying lens formula for eyepiece lens,

 1/ve-1/ue=1/fe

 ⇒1/-25-1/ue=1/6.25

 ⇒ue=-5cm

  • Image distance from objective lens is v0=L-|ue|=15-5=10cm
  • Now applying lens formula for objective lens,

1/vo-1/uo=1/fo

 ⇒1/10-1/uo=1/2

 ⇒uo=-2.5cm

objective lens

Image distance for the eyepiece, ve=-∞

Object distance for the eyepiece, is ue

Applying lens formula for eyepiece lens,

 1/ve-1/ue=1/fe

 ⇒1/-∞-1/ue=1/6.25

 ⇒ue=-6.25cm

objective lens

 

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