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Q.

A telescope has an objective of focal length 50 cm and an eye piece of focal length 5 cm. The least distance of distinct vision is 25 cm. The telescope is focussed for distinct vision on a scale 200 cm away. The separation between the objective and the eye-piece is

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a

75 cm

b

60 cm

c

71 cm

d

74 cm

answer is C.

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Detailed Solution

$f_{o} = 50 cm, f_{θ} = 5 cm, D = 25 cm and u_{o} = 200 cm Separation between the objective and the eye lens is L = \frac{u_{o} f_{o}}{(u_{o} - f_{o})} + \frac{f_{e} D}{(f_{e} + D)} = \frac{200 \times 50}{(200 - 50)} + \frac{5 \times 25}{(5 + 25)} = 71 cm$

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