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Question
An angular magnification (magnifying power) of 30X is desired using an objective of focal length 1.25 cm and an eyepiece of focal length 5 cm. How will you set up the compound microscope?
Solution
Focal length of the objective lens, fo = 1.25 cm
Focal length of the eyepiece, fe = 5 cm
Least distance of distinct vision, d = 25 cm
Angular magnification of the compound microscope = 30X
Total magnifying power of the compound microscope, m = 30
The angular magnification of the eyepiece is given by the relation:
`"m"_"e" = (1 + "d"/"f"_"e")`
= `(1 + 25/5)`
= 6
The angular magnification of the objective lens (mo) is related to me as:
mo me = m
mo = `"m"/"m"_"e"`
= `30/6`
= 5
We also have the relation:
`"m"_"o" = ("Image distance for the objective lens" ("v"_"o"))/("Object distance for the objective lens" (-"u"_"o"))`
`5 = "v"_"o"/-"u"_"o"`
∴ vo = −5uo ................(1)
Applying the lens formula for the objective lens:
`1/"f"_"o" = 1/"v"_"o" - 1/"u"_"o"`
`1/1.25 = 1/(-5"u"_"o") - 1/"u"_"o"`
`1/1.25 = (-6)/(5"u"_"o")`
uo = `(-6)/5 xx 1.25`
∴ uo = −1.5 cm
And vo = −5uo
= −5 × (−1.5)
vo = 7.5 cm
The object should be placed 1.5 cm away from the objective lens to obtain the desired magnification.
Applying the lens formula for the eyepiece:
`1/"v"_"e" - 1/"u"_"e" = 1/"f"_"e"`
Where,
ve = Image distance for the eyepiece = −d = −25 cm
ue = Object distance for the eyepiece
`1/"u"_"e" = 1/"v"_"e" - 1/"f"_"e"`
= `(-1)/25 - 1/5`
= `- 6/25`
∴ ue = −4.17 cm
Separation between the objective lens and the eyepiece = `|"u"_"e"| + |"v"_"o"|`
= 4.17 + 7.5
= 11.67
Therefore, the separation between the objective lens and the eyepiece should be 11.67 cm.
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