An astronomical telescope has objective of focal length 50 cm. For normal vision, its magnification is 5. Now length of the telescope tube is adjusted such that the final image is formed at least distance of distinct vision (D = 25 cm). In this case the magnification produced by the telescope is
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answer is 4.
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Detailed Solution
For normal vision, f0fe=5⇒fe=505cm=10 cmWhen final image is formed at least distance of distinct vision, m=f0fe1+feD=50101+1025=7