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 :

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​