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The focal lengths of the objective and eye lens of a microscope are 1cm and 5 cm respectively. If magnifying power for the relaxed eye is 45, then length of the tube is ____cm.

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answer is 15.

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Detailed Solution

$f_{0} = 1 c m f_{c} = 5 c m m = 45 m_{e} = \frac{D}{f_{e}} = \frac{25}{5} = 5 = m = m_{0} \times m_{e} \Rightarrow 45 = \frac{V_{0}}{U_{0}} \times 5 \Rightarrow V_{0} = 9 U_{0}$

For objective $\frac{1}{f_{0}} = \frac{1}{V_{0}} - \frac{1}{U_{0}}$ $\frac{1}{1} = \frac{1}{V_{0}} - \frac{9}{- V_{0}}$

$V_{0} = 10 c m \Rightarrow L = V_{0} + U_{e} = V_{0} + f_{e} = 10 + 5 = 15 c m$

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