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Question

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.

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Solution

Step 1: Draw diagram of compound microscope.


Step 2: Find the object distance for the eyepiece.

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=25cm

Using lens formula for the eyepiece,

1ve1ue=1fe
1ue=1ve1fe
1ue=12512.5
1ue=1125cm
ue=2.27cm


Step 3: Find the image distance for the objective lens.

Using lens formula for the eyepiece,
1vo1uo=1fo
1vo=1uo+1fo
1vo=10.9+10.8
1vo=0.8+0.90.72=0.10.72
vo=7.2cm


Step 4: Find the distance between the objective lens and the eyepiece.

Distance between objective lens and eye piece is given by
d=vo+|ue|
d=7.2+2.27
d=9.47cm


Step 5: Find the Magnification power of microscope.

Magnification of microscope at the least distance of distinct vision is given by,
m=vo|uo|(1+Dfe)
m=7.20.9(1+252.5)
m=8×11
m=88.

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