In a compound microscope, the magnified virtual image is formed at a distance of 25 cm from the eye-piece. The focal length of its objective lens is 1 cm. If the magnification is 100 and the tube length of the microscope is 20 cm, then the focal length of the eye-piece lens (in cm) is __________. (Round off to the nearest integer)

Solution

Physics sir · 3 steps

Image at 25 cm means near-point viewing, so the eyepiece magnifies by 1 + D/fₑ, not just D/fₑ.

1. Two magnifications. The total magnification is the objective's magnification times the eyepiece's. The objective gives about L/fₒ, where L is the tube length. The final image sits at D = 25 cm, so the eyepiece gives 1 + D/fₑ.

   m = L/fₒ(1 + D/fₑ)

2. Put in numbers. 100 = 20/1(1 + 25/fₑ), so 1 + 25/fₑ = 5. That gives 25/fₑ = 4, so fₑ = 6.25 cm.
3. So. The focal length of the eyepiece is 6.25 cm, which rounds to 6.

The common slip: 5. Using the relaxed-eye formula m = L/fₒ·D/fₑ gives fₑ = 5 cm. The image is at 25 cm, so the eyepiece term must be 1 + D/fₑ.

The idea to keep. Check where the final image is. At the near point (25 cm) the eyepiece gives 1 + D/fₑ. At infinity (relaxed eye) it gives D/fₑ. Mixing them up shifts your answer.

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