A personwith a normal near point (25 cm) using a compound microscope withobjective of focal length 8.0 mm and an eyepiece of focal length 2.5cm can bring an object placed at 9.0 mm from the objective in sharpfocus. What is the separation between the two lenses? Calculate themagnifying power of the microscope,
Focallength of the objective lens, fo= 8 mm = 0.8 cm
Focallength of the eyepiece, fe= 2.5 cm
Objectdistance for the objective lens, uo= −9.0 mm = −0.9 cm
Leastdistance of distant vision, d =25 cm
Imagedistance for the eyepiece, ve= −d= −25 cm
Objectdistance for the eyepiece =
Usingthe lens formula, we can obtain the value ofas:
Wecan also obtain the value of the image distance for the objectivelensusing the lens formula.
Thedistance between the objective lens and the eyepiece
Themagnifying power of the microscope is calculated as:
Hence, themagnifying power of the microscope is 88.