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Question
Solve Numerical example.
Focal power of the eye lens of a compound microscope is 6 dioptre. The microscope is to be used for maximum magnifying power (angular magnification) of at least 12.5. The packing instructions demand that length of the microscope should be 25 cm. Determine minimum focal power of the objective. How much will its radius of curvature be if it is a biconvex lens of n = 1.5?
Solution
Focal power of the eye lens,
Pe = `1/"f"_"e"` = 6 D
∴ fe = `1/6` = 0.1667 m = 16.67 cm
Now, as the magnifying power is maximum,
ve = 25 cm,
Also, (Me)max = `1+"D"/"f"_"e"=1+25/16.67` ≈ 2.5
Given that,
M = mo × Me = 12.5
∴mo × 2.5 = 12.5
∴ mo = `"v"_"o"/"u"_"o"` = 5 .......(i)
From thin lens formula,
`1/"f"_"e"=1/"v"_"e"-1/"u"_"e"`
∴ `1/16.67=1/-25-1/"u"_"e"`
∴ `1/"u"_"e"=1/-25-1/16.67`
∴ ue ≈ −10 cm .......(ii)
Length of a compound microscope,
L = |vo| + |ue|
∴ 25 = |vo| + 10
∴ |vo| = 15 cm
∴ |uo|= `"v"_"o"/5` = 3 cm ......[from (i)]
From lens formula for objective,
`1/"f"_"o"=1/"v"_"o"-1/"u"_"o"`
= `1/15-1/-3`
= `2/5`
∴ fo = 2.5 cm = 0.025 m
Thus, focal power of objective,
P = `1/("f"_"o"("m"))`
= `1/0.025` = 40 D
Using lens maker’s equation for a biconvex lens,
`1/"f"_"o"=("n"-1)(1/"R"-1/(-"R"))`
∴ `1/2.5=(1.5-1)(2/"R")=1/"R"`
∴ R = 2.5 cm
The focal power and radius of curvature of objective lens are 40 D and 2.5 cm respectively.
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