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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
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4
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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