For a magnification of 375 from a compound microscope of tube length 15 cm and an objective of focal length 5 mm, the focal length of the eye-piece, should be close to:

Options

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

Answer

D

Explanation

Case-I
If final image is at least distance of clear vision.
`M.P = L/f_(0) (1 + D/f_(e)), 375 = 150/5 [1 + 25/f_(e)]` `375/30 = 1 + 25/f_(e)`
`345/30 = 25/f_(e)`
`f_(e) = 750/345 = 2.17 cm, f_(e) ~~ 22 mm`
Case-ll
If final image is at infinity
M.P =` L/(f_(0)) = (D/f_(e)) = 375`
`f_(e) = 22 mm`