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An equiconvex lens of focal length 10 cm and $μ = 1.5$ cut into two identical pieces and these two places are placed in contact with each other as shown in figure. The spaced between the plano-convex lenses is filled with a liquid of refractive index 4/3. Then the equivalent focal length of the combination of three lenses is

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20 cm

12 cm

7.5 cm

15 cm

answer is B.

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Detailed Solution

$\frac{1}{f} = \frac{2 (μ - 1)}{R} \Rightarrow \frac{1}{10} = \frac{2 (1.5 - 1)}{R} \Rightarrow R = 10 c m$

Now, $f_{1} =$ focal length of each plano convex lens

$= \frac{R}{(μ - 1)} = \frac{10}{(1.5 - 1)} c m = 20 c m$

and focal length of the liquid lens is

$\frac{1}{f_{2}} = - \frac{2 (4 / 3 - 1)}{10} \Rightarrow f_{2} = - 15 c m$

$∴ \frac{1}{F} = \frac{1}{f_{1}} + \frac{1}{f_{2}} + \frac{1}{f_{3}} = \frac{1}{20} + \frac{1}{(- 15)} + \frac{1}{20} \Rightarrow F = 30 c m$

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