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Q.

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

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a

12 cm

b

25 cm

c

15 cm

d

30 cm

answer is C.

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Detailed Solution

$\begin{array}{l} m = m_{0} m_{e}, m_{e} = \frac{D}{f_{e}} = 5 \\ ∴ m_{0} = \frac{45}{5} = 9 = \frac{v_{0}}{u_{0}}; u_{0} = \frac{v_{0}}{9} \\ \frac{1}{f_{0}} = \frac{1}{v_{0}} + \frac{1}{u_{0}}; \frac{1}{1} = \frac{1}{v_{0}} + \frac{9}{v_{0}} = \frac{10}{v_{0}} \\ v_{0} = 10 cm; L = v_{0} + f_{e} = 10 + 5 = 15 cm \end{array}$

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