In a compound microscope, the focal lengths of two lenses are 1.5 cm and 6.25 cm an object is placed at 2 cm from objective and the final image is formed at 25 cm from the eye lens. The distance between the two lenses is (in cm)

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9.25

7.75

answer is D.

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Detailed Solution

$f_{0} = 1.5 c m, f_{c} = 6.25 c m; u_{0} = - 2 c m V_{e} = - 25 c m$
for eye piece $\frac{1}{f_{c}} = \frac{1}{V_{e}} - \frac{1}{U_{e}}; \frac{4}{25} = \frac{1}{- 25} - \frac{1}{U_{e}}$
$U_{e} = - 5 c m$
For eye piece $\frac{1}{f_{0}} = \frac{1}{V_{0}} - \frac{1}{U_{0}}; \frac{2}{3} = \frac{1}{V_{0}} - \frac{1}{- 2}$
$\frac{2}{3} - \frac{1}{2} = \frac{1}{V_{0}}; \frac{1}{V_{0}} = \frac{4 - 3}{6} V_{0} = 6 cm; L = V_{0} + U_{e} = 11 cm$