A person with a normal near point (25 cm) using a compound microscope with objective of focal length 8.0 mm.

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asked Jan 6, 2018 in Physics by sforrest072 (157,439 points) 63 448 1279
edited Jan 6, 2018 by sforrest072

A person with a normal near point (25 cm) using a compound microscope with objective of focal length 8.0 mm and an eyepiece of focal length 2.5 cm can bring an object placed at 9.0 mm from the objective in sharp focus. What is the separation between the two lenses? Calculate the magnifying power of the microscope,

1 Answer

+1 vote
answered Jan 6, 2018 by mdsamim (213,225 points) 5 10 22
edited Mar 5, 2018 by Vikash Kumar
 
Best answer

Focal length of the objective lens, fo = 8 mm = 0.8 cm 

Focal length of the eyepiece, fe = 2.5 cm

Object distance for the objective lens, uo = −9.0 mm = −0.9 cm 

Least distance of distant vision, d = 25 cm 

Image distance for the eyepiece, ve = −d = −25 cm

Object distance for the eyepiece =uo

 Using the lens formula, we can obtain the value of uoas:

Hence, the magnifying power of the microscope is 88.

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