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If $f_{V}$ and $f_{R}$ are the magnitude of focal lengths of a convex lens for violet and red light respectively and $F_{V}$ and $f_{R}$ are the magnitude of focal lengths of concave lens for violet and red light respectively, then we have

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$f_{V} < f_{R} a n d F_{V} < F_{R}$

$f_{V} < f_{R} a n d F_{V} > F_{R}$

$f_{V} > f_{R} a n d F_{V} > F_{R}$

$f_{V} > f_{R} a n d F_{V} < F_{R}$

answer is B.

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Detailed Solution

$\frac{1}{f} = (μ - 1) (\frac{1}{R_{1}} - \frac{1}{R_{2}})$

According to Cauchy relation $μ = A + \frac{B}{λ^{2}} + \frac{C}{λ^{4}} ......$

Hence $f \propto λ$ .

For violet, wavelength is less hence its focal length is also less. Hence , focus of violet is closer to lens compared to red.

$∴ f_{v} < f_{r}$ and $F_{v} < F_{r}$

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