If we need a magnification of $$375$$ from a compound microscope of tube length $$150$$ mm and an objective of focal length $$5mm$$, the focal length of the $$eye-piece$$, should be close to :

Question

Infinity Learn · Accepted Answer

$$Case-IIf$$ final image is at least distance of clear visionM.P.$$=Lf01+Dfe$$​⇒$$375=15051+25fe$$         ​⇒$$37530=1+25fe$$             ​⇒$$34530=25fe$$     ⇒$$fe=750345=2.17cm$$​⇒fe≈$$22mm$$ $$Case-IIIf$$ final image is at infinityM.$$P=Lf0Dfe=375$$ ​⇒$$37530=25fe$$    ​⇒  fe $$=$$ $$750375$$                         ​ ⇒$$fe=2$$ cm​⇒fe≈$$22$$ mm​

If we need a magnification of 375 from a compound microscope of tube length 150 mm and an objective of focal length 5mm, the focal length of the eye-piece, should be close to :

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