A telescope has an objective of focal length 50 cm and eyepiece of focal length 5 cm. The least distance of distinct vision is 25 cm. The telescope is focused for distinct vision on a scale 200 cm away from the objective. Calculate

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Detailed Solution

$\begin{array}{l} \frac{1}{v_{0}} + \frac{1}{200} = \frac{1}{50} \\ v_{0} = \frac{200}{3} \\ \frac{1}{ue} = \frac{1}{25} + \frac{1}{5} ue = \frac{25}{6} \\ ℓ = \frac{425}{6} = 70.83 cm \end{array}$

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