In a compound microscope the objective and eyepiece have focal length of 0.95cm and 5cm and are kept at a distance of 20cm. If the final image is formed at least distance of distant vision, the position of object and the total magnification is

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$\frac{95}{94} cm, 94$

0.95cm, 95

0.94cm, 94

$\frac{94}{95} cm, 95$

answer is A.

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Detailed Solution

$\begin{array}{l} f_{0} = 0.95 cm, fe = 5 cm \\ L = V_{0} + U_{e} = 20 cm, V_{e} = - 25 ccm \end{array}$

for eyepiece
$\begin{array}{l} \frac{1}{f_{e}} = \frac{1}{v_{e}} - \frac{1}{u_{e}}; \frac{1}{5} = \frac{1}{- 25} - \frac{1}{u_{e}}; u_{e} = \frac{- 25}{6} cm \\ v_{0} + u_{e} = 20; v_{0} = 20 = \frac{25}{6} = \frac{95}{6} \\ \frac{1}{f_{0}} = \frac{1}{v_{0}} = - \frac{1}{u_{0}}; \frac{100}{95} = \frac{6}{95} - \frac{1}{u_{0}}; u_{0} = \frac{95}{94} cm \\ m = \frac{v_{0}}{u_{0}} \times \frac{v_{e}}{u_{e}} = \frac{95 / 6}{\frac{95}{94}} \times \frac{25}{(\frac{25}{6})} = 94 \end{array}$