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Q.

The focal length of objective and the eyepiece of a compound microscope are 1cm and 5cm. An object is placed at a distance of 1.1cm from the objective and if the final image is formed at infinity the magnifying power and distance between the lenses are respectively

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a

50, 15.7cm

b

50, 16cm

c

60, 16.7cm

d

60, 16cm

answer is B.

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Detailed Solution

$\begin{array}{l} \Rightarrow \frac{1}{f_{0}} = \frac{1}{v_{0}} + \frac{1}{μ_{0}} \\ ∴ \frac{1}{1} = \frac{1}{v_{0}} + \frac{1}{1 .1} \Rightarrow v_{0} = 11 cm \\ \Rightarrow m = \frac{v_{0}}{μ_{0}} (\frac{D}{f_{e}}) = \frac{11}{1 .1} (\frac{25}{5}) = 50 \\ \Rightarrow L = v_{0} + f_{e} = 11 + 5 = 16 cm \end{array}$

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