Ray Optics

JEE Main level, single correct, medium

283 of 295

Question 283 of 295

If we need a magnification of 375375 from a compound microscope of tube length 150mm150mm and an objective of focal length 5mm,5mm, the focal length of the eye-piece, should be close to:

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Answer: (A) 22mm22mm

M.P.=Lf0(1+Dfe);375=1505[1+25fe]M.P.=\dfrac{L}{{f}_{0}}\left(1+\dfrac{D}{{f}_{e}}\right);375=\dfrac{150}{5}\left[1+\dfrac{25}{{f}_{e}}\right]
37530=1+25fe\dfrac{375}{30}=1+\dfrac{25}{{f}_{e}}
34530=25fe\dfrac{345}{30}=\dfrac{25}{{f}_{e}}
fe=750345=2.17cm;fe≈22mm{f}_{e}=\dfrac{750}{345}=2.17cm;{f}_{e}\approx 22mm

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