Question

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 \[ (\text{Least distance of distinct vision}=25\,\text{cm}) \]

• A. \(24\)
• B. \(15\)
• C. \(12\)
• D. \(20\)

Hint:

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.

Answer 1

The Correct Option is D

Step 1: Recall the magnification formulas. For a compound microscope, - When the final image is at infinity, \[ M_\infty = \frac{L}{f_o}\cdot\frac{D}{f_e}. \] - When the final image is at the near point, \[ M_N = \frac{L}{f_o} \left(1+\frac{D}{f_e}\right). \] Hence, \[ M_N-M_\infty = \frac{L}{f_o}. \]

Step 2: Substitute the given values. Given, \[ L=100\,\text{cm}, \] \[ f_o=5\,\text{cm}. \] Therefore, \[ M_N-M_\infty = \frac{100}{5} = 20. \] Thus, \[ \boxed{\text{Difference in magnification}=20.} \] Hence, the correct option is \(\boxed{(D)}\).