In the diagram given below, there are three lenses formed. Considering negligible thickness of each of them as compared to $|R_{1}|$ and $|R_{2}|$ , i.e., the radii of curvature for upper and lower surfaces of the glass lens, the power of the combination is

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$\frac{1}{6} (\frac{1}{|R_{1}|} + \frac{1}{|R_{2}|})$

$\frac{1}{6} (\frac{1}{|R_{1}|} - \frac{1}{|R_{2}|})$

$- \frac{1}{6} (\frac{1}{|R_{1}|} + \frac{1}{|R_{2}|})$

$- \frac{1}{6} (\frac{1}{|R_{1}|} - \frac{1}{|R_{2}|})$

answer is B.

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Detailed Solution

$\begin{array}{l} \Rightarrow p_{e} = p_{1} + p_{2} + p_{3} \\ Power of 1st lens, p_{1} = (\frac{4}{3} - 1) (\frac{1}{\infty} - \frac{1}{- |R_{1}|}) \\ \Rightarrow p_{1} = (\frac{1}{3 |R_{1}|}) \\ Power of 2nd lens, p_{2} = (\frac{1}{2}) (\frac{1}{- |R_{1}|} - \frac{1}{- |R_{2}|}) \\ \Rightarrow p_{2} = \frac{1}{2} (\frac{1}{|R_{2}|} - \frac{1}{|R_{1}|}) \\ Power of 3rd lens, p_{3} = (\frac{1}{3}) (\frac{1}{- |R_{2}|} - \frac{1}{\infty}) = - \frac{1}{3 |R_{2}|} \\ Thus, p_{eq} = \frac{1}{3} (\frac{1}{|R_{1}|} - \frac{1}{|R_{2}|}) - \frac{1}{2} (\frac{1}{|R_{1}|} - \frac{1}{|R_{2}|}) \\ = - \frac{1}{6} (\frac{1}{|R_{1}|} - \frac{1}{|R_{2}|}) \end{array}$