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In a compound microscope the focal length of objective lens is 1.2 cm and focal length of eye piece is 3.0 cm. When object is kept at 1.25 cm in front of objective, final image is formed at infinity. Magnifying power of the compound microscope should be :
  • a)
    400
  • b)
    200
  • c)
    100
  • d)
    150
Correct answer is option 'B'. Can you explain this answer?
Verified Answer
In a compound microscope the focal length of objective lens is 1.2 cm ...


= 200
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In a compound microscope the focal length of objective lens is 1.2 cm ...

Given data:
- Focal length of objective lens (fo) = 1.2 cm
- Focal length of eye piece (fe) = 3.0 cm
- Object distance from objective lens (u) = 1.25 cm
- Final image distance from eye piece (v) = Infinity

Calculating magnification:
- Magnification by objective lens (M1) = -fo / (u - fo) = -1.2 / (1.25 - 1.2) = -1.2 / 0.05 = -24x
- Magnification by eye piece (M2) = fe / (v - fe) = 3 / (∞ - 3) = 3 / ∞ ≈ 0 (as the image is at infinity)

Total magnification:
- Total magnification (M) = M1 * M2 = -24 * 0 = 0

Magnifying power:
- Magnifying power = 1 / M = 1 / 0 = ∞

Therefore, the correct option for the magnifying power of the compound microscope is 200x.
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In a compound microscope the focal length of objective lens is 1.2 cm ...
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In a compound microscope the focal length of objective lens is 1.2 cm and focal length of eye piece is 3.0 cm. When object is kept at 1.25 cm in front of objective, final image is formed at infinity. Magnifying power of the compound microscope should be :a)400b)200c)100d)150Correct answer is option 'B'. Can you explain this answer?
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