The objective and eye piece of an astronomical telescope are double convex lenses with refractive index 1.5. When the telescope is adjusted to infinity the separation between the lenses is 16 cm. If the space between the lenses is now filled with water and again telescope is adjusted for infinity. Then the present separation between the lenses is

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8 cm

16 cm

24 cm

32 cm

answer is D.

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Detailed Solution

$\begin{array}{l} f_{0} = \frac{R_{0}}{2 (μ - 1)} = \frac{R_{0}}{2 (1 .5 - 1)} = R_{0} \\ f_{e} = \frac{R_{e}}{2 (μ - 1)} = R_{e}; L = f_{0} + f_{e} = R_{0} + R_{e} = 16 \end{array}$
When water in between the lenses for objective
1st surface $\frac{1.5}{v} - \frac{1}{- \infty} = \frac{1.5 - 1}{R_{0}}$
2nd surface $\frac{4}{3 v_{0}} - \frac{1.5}{V} = \frac{\frac{4}{3} - 1.5}{- R_{0}}$
$\Rightarrow \frac{4}{3 v_{0}} = \frac{2}{3 R_{0}} \Rightarrow v_{0} = 2 R_{0}$
Similarly $u_{e} = 2 R_{e}$
$L^{1} = v_{0} + u_{e} = 2 (R_{0} + R_{e}) = 2 \times 16 = 32 cm$