Ray Optics

JEE Main level, single correct, medium

282 of 295

Question 282 of 295

The magnifying power of a telescope with tube length 60cm60cm is 5.5. What is the focal length of its eye piece?

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Answer: (D) 10cm10cm

m=f0fem=\dfrac{{f}_{0}}{{f}_{e}}
5=f0fe5=\dfrac{{f}_{0}}{{f}_{e}}
fo=5fe{f}_{o}=5{f}_{e}
fo+fe=60{f}_{o}+{f}_{e}=60
6fe=606{f}_{e}=60
fe=10{f}_{e}=10

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