An equi-convex lens $(μ = 1.5)$ is combined with an equi concave lens $(μ = 1.3)$ . The radii of curvature of all surface is the same. (assume thin lenses)

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The combination behaves like a converging lens in medium of refractive index 1.4 and diverging lens in a medium of refractive index 1.6.

The combination behaves like a converging lens in a medium of refractive index 1.2 as well as 1.4.

The combination behaves like a diverging lens in medium of refractive index 1.2 as well as 1.6.

The combination behaves like a diverging lens in medium of refractive index 1.4 and converging lens in a medium of refractive index 1.6.

answer is B.

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Detailed Solution

$\begin{array}{l} F_{e q} \Rightarrow + v e c o n v e r g i n g \\ F_{e q} \Rightarrow - v e d i v e r g i n g \end{array}$

$\frac{1}{F_{e q}} = \frac{1}{F_{1}} + \frac{1}{F_{2}}$

$\frac{1}{F_{1}} = [\frac{μ_{1}}{μ_{m}} - 1] [\frac{2}{R}]$

$a n d \frac{1}{F_{2}} = [\frac{μ_{2}}{μ_{m}} - 1] [\frac{- 2}{R}]$

$\frac{1}{F_{e q}} = \frac{2}{R} [(\frac{μ_{1}}{μ_{m}} - 1) - (\frac{μ_{2}}{μ_{m}} - 1)]$

$\frac{1}{F_{e q}} = \frac{2}{R} [\frac{μ_{1}}{μ_{m}} - \frac{μ_{2}}{μ_{m}}]$

(A) $\frac{1}{F_{e q}} = \frac{2}{R} [\frac{1.5}{1.4} - \frac{1.3}{1.4}],$ Converging

$\frac{1}{F_{e q}} = \frac{2}{R} [\frac{1.5}{1.6} - \frac{1.3}{1.6}]$ , Converging

(B) $\frac{1}{F_{e q}} = \frac{2}{R} [\frac{1.5}{1.2} - \frac{1.3}{1.2}],$ Converging

$\frac{1}{F_{e q}} = \frac{2}{R} [\frac{1.5}{1.4} - \frac{1.3}{1.4}]$ , Converging
(C) $\frac{1}{F_{e q}} = \frac{2}{R} [\frac{1.5}{1.6} - \frac{1.3}{1.6}],$ Converging
$\frac{1}{F_{e q}} = \frac{2}{R} [\frac{1.5}{1.2} - \frac{1.3}{1.2}]$ Converging