MHT CET · Physics · Ray Optics
A compound microscope produces a magnification of 24 . The focal length of the eyepiece is \(5 \mathrm{~cm}\). The final image is formed at the least distance of distinct vision. The magnification produced by the objective is
- A 4
- B 5
- C 6
- D 7
Answer & Solution
Correct Answer
(A) 4
Step-by-step Solution
Detailed explanation
For a compound microscope, \(M=m_o \times m_e\) and since the final image is formed at \(D=25 \mathrm{~cm}\),
\(\begin{aligned} & M=m_o\left(1+\frac{D}{f_e}\right) \\ & \therefore m_o=\frac{M}{1+\frac{D}{f_e}}=\frac{24}{1+\frac{25}{5}} \\ & \therefore m_e=\frac{24}{1+5}=4\end{aligned}\)
\(\begin{aligned} & M=m_o\left(1+\frac{D}{f_e}\right) \\ & \therefore m_o=\frac{M}{1+\frac{D}{f_e}}=\frac{24}{1+\frac{25}{5}} \\ & \therefore m_e=\frac{24}{1+5}=4\end{aligned}\)
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