A compound microscope has a magnifying power of 100 when the image is formed at infinity. The objective has a focal length of 0.5 cm and the tube length is 6.5 cm. Find the focal length of the eyepiece.
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4
Given :
Magnifying power, m = 100
Focal length of the objective, fo = 0.5 cm
Tube length, l = 6.5 cm
Since the image is formed at infinity, the real image produced by the objective lens should lie on the focus of the eyepiece.
∴ vo + fe = 6.5 cm …(1)
The magnifying power for normal adjustment is given by
m=[vo/uo]×D /fe
= -[1-vo/fo] D/fe
⇒100 = -[1-vo/0.5]×25/fe
⇒2vo-4fe=1 …2
On solving equations (1) and (2), we get:
Vo = 4.5 cm and fe = 2 cm
Thus, the focal length of the eyepiece is 2 cm.
Answered by
0
Answer:
The magnifying power of a compound microscope is given by:
m=
f
o
f
e
LD
⇒f
e
=
f
o
m
LD
=
0.5×100
6.5×2.5
Substituting the respective values, we get f
e
=3.25
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