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Question
Refractive index of a flint glass varies from 1.60 to 1.66 for visible range. Radii of curvature of a thin convex lens are 10 cm and 15 cm. Calculate the chromatic aberration between extreme colours.
Solution
Given the refractive indices for extreme colours.
As, nR < nV
nR = 1.60 and nV = 1.66
For convex lens,
R1 = 10 cm and R2 = – 15 cm
∴ `(1/"R"_1-1/"R"_2)=(1/10-1/-15)=1/6`
For red colour,
`1/"f"_"R"=("n"_"R"-1)(1/"R"_1-1/"R"_2)`
= (1.60 − 1)`1/6`
= 0.1
∴ fR = 10 cm
Similarly, for violet colour,
`1/"f"_"V"=("n"_"V"-1)(1/"R"_1-1/"R"_2)`
= (1.66 − 1)`1/6`
= `0.66 xx 1/6`
= `0.66/6`
`=> 1/"f"_"V"` = 0.11
`=> "f"_"V" = 1/0.11`
`=> "f"_"V" = 1/(11 xx 10^-1)`
`=> "f"_"V" = 1/1.1 xx 10^1`
∴ fV = 0.9091 × 101
∴ fV = 9.091 cm
∴ Longitudinal chromatic aberration -
= fR − fV
= 10 − 9.091
= 0.909 cm ≈ 0.91 cm
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