A thin convex lens of focal length 10cm and refractive index 1.5 is cut vertically into two equal pieces. They are placed as shown with a liquid of refractive index 3 between them. What is the focal length of the combination?

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-10/4cm

-10/3cm

None

-10cm

answer is C.

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Detailed Solution

$\begin{array}{l} \frac{1}{f} = (μ - 1) (\frac{1}{R_{1}} - \frac{1}{R_{2}}) \\ \Rightarrow R = 10 cm \\ f^{'} = (3 - 1) (\frac{1}{- 10} - \frac{1}{10}) \\ \Rightarrow f^{'} = - \frac{10}{4} \\ \frac{1}{f_{eq}} = \frac{1}{20} - \frac{4}{10} + \frac{1}{20} = \frac{2}{20} - \frac{4}{10} \\ \Rightarrow f_{eq} = - \frac{10}{3} cm \end{array}$

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