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Question
An angular magnification of 30X is desired using an objective of focal length 1.25 cm and an eye piece of focal length 5 cm. How will you set up the compound microscope for the final image formed at least distance of distinct vision?
Solution
mo = 30, fo = 1.25 cm, fe = 5 cm
When image is formed at least distance of distinct vision,
D = 25 cm
Angular magnification of eyepiece
`"m"_"e" = (1 + "D"/"f"_"e") = 1 + 25/5 = 6`
Total Angular magnification,
`"m" = "m"_"o" "m"_"e" => "m"_"o" = "m"/"m"_"e" = 30/6 = 5`
As the objective lens forms the real image,
`"m"_"o" = "v"_"o"/"u"_"o" = -5 => "v"_"o" = -5"u"_"o"`
using lens equation,
uo = −1.5 cm, vo = −5 × (−1.5) cm = +7.5 cm
Given ve = −D = −25 cm, fe = +5 cm, ue = ?
using again lens equation ue = `25/6`
Thus, the object is to be placed at 1.5 cm from the objective and separation between the two lenses should be
L = vo + Iue l = 11.67 cm
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