JEE Main20207 Jan 2020Morning ShiftPhysicsRay OpticsActual
If we need a magnification of 375 from a compound microscope of tube length 150 m m and an objective of focal length 5 m m , the focal length of the eye-piece, should be close to:
Options
- A22 m m
- B2 m m
- C4 m m
- D33 m m
Correct answer
A. 22 m m
Step-by-step solution
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 c m ; f e ≈ 22 m m