TS EAMCET · Physics · Ray Optics
The total magnification produced by a compound microscope is 24 when the final image is formed at the least distance of distinct vision. If the focal length of the eyepiece is 5 cm, the magnification produced by the objective is
- A \(4\)
- B \(4.8\)
- C \(120\)
- D \(6\)
Answer & Solution
Correct Answer
(A) \(4\)
Step-by-step Solution
Detailed explanation
\(M_o = \frac{M}{1 + \frac{D}{f_e}}\) \(M_o = \frac{24}{1 + \frac{25 \text{ cm}}{5 \text{ cm}}}\) \(M_o = \frac{24}{1 + 5}\) \(M_o = \frac{24}{6}\) \(M_o = 4\)
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