A plano convex lens of refractive index $μ_{1}$ and focal length $f_{1}$ is kept in contact with another plano concave lens of refractive index $μ_{1}$ and focal length $f_{2} .$ If the radius of curvature of their spherical faces is R each and $| f_{1} | = 2 | f_{2} ∣$ , then $μ_{1} & μ_{2}$ are related as:

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$μ_{1} + μ_{2} = 3$

$2 μ_{1} - μ_{2} = 1$

$3 μ_{2} - 2 μ_{1} = 1$

$2 μ_{2} - μ_{1} = 1$

answer is B.

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Detailed Solution

From lens maker’s formula,

$\frac{1}{f} = (μ - 1) (\frac{1}{R_{1}} - \frac{1}{R_{2}}); \frac{1}{f_{1}} = (μ_{1} - 1) (\frac{1}{\infty} - \frac{1}{- R}) = \frac{μ_{1} - 1}{R}$

Similarly, for the other plano-concave lens

$\frac{1}{f_{2}} = (μ_{2} - 1) (\frac{1}{- R} - \frac{1}{\infty}) = \frac{- (μ_{2} - 1)}{R}$

from the relation of magnitude of focal lengths we get

$\frac{(μ_{1} - 1)}{R} = \frac{(μ_{2} - 1)}{2 R} \Rightarrow 2 μ_{1} - μ_{2} = 1$

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