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

Options

A. 2 mm
B. 22 mm
C. 12 mm
D. 33 mm

Answer

A, B

Explanation

Case-I:
If final image is obtained at near point, then magnification,
`m=(L)/(f_0)(1-(D)/(f_e))`
`375=(150)/(50)(1+(25)/(f_e))`
`(375)/(3)=1+(25)/(f_e)`
`123-1=(25)/(f_e)`
`f_e=(25)/(122)`
`thereforef_e=0.2049`mm `therefore~~0.2` mm
Case-II :
If final image is obtained at infinity, then magnification,
`m=(L)/(f_0)((D)/(f_e))`
`375=(150)/(50)((25)/(f_e))`
`(375)/(3)=(25)/(f_e)`
`thereforef_e=(25xx3)/(375)` `thereforef_e=0.2` mm