A simple telescope consisting of an objective lens of focal length 30 cm and that of its eye-lens is 3 cm. It is focused on a scale 2 metres distant from it. The distance of objective lens from eye-lens to see with relaxed eye is

33 cm

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65.3 cm

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38.3 cm

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40.3 cm

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Solution

The correct option is C
38.3 cm

$u_{0} = - 200 cm; f_{0} = 30 cm$

$\frac{1}{v_{0}} - \frac{1}{- 200} = \frac{1}{+ 30} \Rightarrow v_{0} = 35.3 cm$

$L = v_{0} + f_{e} = 35.3 + 3 = 38.3 cm$

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A simple telescope consisting of an objective lens of focal length 30 cm and that of its eye-lens is 3 cm. It is focused on a scale 2 metres distant from it. The distance of objective lens from eye-lens to see with relaxed eye is

The correct option is C 38.3 cm u0=−200 cm; f0=30 cm 1v0−1−200=1+30⇒v0=35.3 cm L=v0+fe=35.3+3=38.3 cm

The correct option is C
38.3 cm

$u_{0} = - 200 cm; f_{0} = 30 cm$

$\frac{1}{v_{0}} - \frac{1}{- 200} = \frac{1}{+ 30} \Rightarrow v_{0} = 35.3 cm$

$L = v_{0} + f_{e} = 35.3 + 3 = 38.3 cm$