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

Solution

Physics sir · 4 steps

Split the 375 into objective times eyepiece: the objective gives 30, so the eyepiece must give 12.5.

1. Two magnifications multiply. A compound microscope's total magnification is the objective's magnification times the eyepiece's. The objective gives about L/fₒ, and the eyepiece gives 1 + D/fₑ when the final image is at the near point, D = 250 mm.

   M = L/fₒ(1 + D/fₑ)

2. Objective part. L/fₒ = 150/5 = 30. So the eyepiece must supply 375/30 = 12.5.
3. Solve for fe. 1 + 250/fₑ = 12.5 gives 250/fₑ = 11.5, so fₑ = 250/11.5 ≈ 21.7 mm.
4. So. The eyepiece focal length is close to 22 mm, option (a).

The idea to keep. For a compound microscope, find the objective's share L/fₒ first. The rest of the magnification belongs to the eyepiece, and 1 + D/fₑ (with D = 25 cm) then gives fₑ.

Answer from the NTA final answer key. Solution by PrepWiser.