An astronomical telescope has an angular magnification of magnitude 5 for distant objects. The seperation between the objective and the eye piece is 36 cm and the final image is formed at infinity. The focal length f0 of the objective and the focal length fe of the eye piece are

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f0 = 45 cm, fe = –9 cm

f0 = 7.2 cm, f = 5 cm

f0 = 50 cm, fe = 10 cm

f0 = 30 cm, fe = 6 cm

answer is D.

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Detailed Solution

m= f0fe⇒5=f0fe------(1) L= f0+fe⇒36=f0+fe----(2) Given From (1) & (2)f0 = 30cm, fe = 6cm

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