An object viewed from a near point distance of 25 cm, using a microscopic lens with magnification '6', gives an unresolved image. A resolved image is observed at infinite distance with a total magnification double the earlier using an eyepiece along with the given lens and a tube of length 0.6 m, if the focal length of the eyepiece is equal to __________ cm.

Solution

Physics sir · 4 steps

First get the lens's focal length from the simple microscope, then use it as the objective of a compound one.

1. Simple microscope. With the image at the near point, m = 1 + D/f. So 6 = 1 + 25/f, which gives f = 5 cm. This lens is now the objective, fₒ = 5 cm.
2. Compound microscope. With the final image at infinity, the total magnification is the objective's magnification times the eyepiece's. Taking the image distance of the objective as the tube length L = 60 cm:

   m = L/fₒ × D/fₑ

3. Put in numbers. The new magnification is double the old one, so m = 12. Then 12 = 60/5 × 25/fₑ = 12 × 25/fₑ.
4. So. 25/fₑ = 1, so the focal length of the eyepiece is 25 cm.

The common slip: 30. Using m = D/f for the simple microscope gives fₒ = 25/6 cm and then fₑ = 30 cm. The image is at the near point here, so the formula needs the extra 1.

The idea to keep. Simple microscope: 1 + D/f for the image at the near point, D/f for the image at infinity. Compound microscope at infinity: L/fₒ×D/fₑ. Read where the image forms before choosing.

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