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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 $μ_{2}$ and focal length $f_{2}$ . If the radius of curvature of their spherical faces is R each and $f_{1} = 2 f_{2}$ , then $μ_{1} a n d μ_{2}$ are related as

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$3 μ_{2} - 2 μ_{1} = 1$

$2 μ_{2} - μ_{1} = 1$

$2 μ_{1} - μ_{2} = 1$

$μ_{1} + μ_{2} = 3$

answer is C.

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Detailed Solution

Given $f_{1} = 2 f_{2}$

$\Rightarrow \frac{1}{| f_{1} |} = \frac{1}{| 2 f_{2} |} \to (1)$

Using lens maker’s formula , weget

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

$\frac{1}{f_{2}} = (μ_{2} - 1) (\frac{1}{{R_{1}}^{1}} - \frac{1}{{R_{2}}^{1}})$

$\Rightarrow | (μ_{1} - 1) (\frac{1}{α} - \frac{1}{- R}) |$

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

(using equation (1)

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

$\Rightarrow 2 μ_{1} - μ_{2} = 1$

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