Question:medium

If the length of a compound microscope is \(100\,\text{cm}\) and the focal length of its objective is \(5\,\text{cm}\), then the difference between the magnifications of the microscope when the final image forms at infinity and at the near point is \[ (\text{Least distance of distinct vision}=25\,\text{cm}) \]

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Remember, \[ \boxed{ M_\infty=\frac{L}{f_o}\cdot\frac{D}{f_e} } \] and \[ \boxed{ M_N=\frac{L}{f_o}\left(1+\frac{D}{f_e}\right). } \] Therefore, \[ \boxed{ M_N-M_\infty=\frac{L}{f_o}, } \] which is independent of the eyepiece focal length.
Updated On: Jul 18, 2026
  • \(24\)
  • \(15\)
  • \(12\)
  • \(20\)
Show Solution

The Correct Option is D

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