A telescope has an objective lens of focal length 150 cm and an eyepiece of focal length 5 cm. If a 50 m tall tower at a distance of 1 km is observed through this telescope in normal setting, the angle formed by the image of the tower is $θ$ , then $θ$ is close to:

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60⁰

1.5 rad

30⁰

1⁰

answer is B.

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Detailed Solution

$\begin{array}{l} u_{0} = 1 km; v_{0} \approx f_{0} \\ \frac{size of object}{size of image} = \frac{u_{0}}{v_{0}} \\ \frac{50}{I} = \frac{1000}{150 \times 10^{- 2}} \\ I = \frac{50 \times 150 \times 10^{- 2}}{1000} m = \frac{15}{2} cm \end{array}$

$\tan θ_{i} = \frac{I}{fe} = \frac{15 / 2}{5} = 1 .5; θ_{i} = \tan^{- 1} (1 .5) \approx 60^{0}$

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For Gallelion telescope for near point adjustment
$m = \frac{f_{0}}{fe} [1 - \frac{fe}{D}] = \frac{30}{3} (1 - \frac{3}{25}) = + 8 .8$

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