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A microscope has an objective of aperture $8 \mathrm{~mm}$ and focal length of $5 \mathrm{~cm}$. The minimum separation between two objects to be just resolved by the microscope is (wavelength of light used $=5500 Å$ )
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Verified Answer
The correct answer is:
$4.2 \mu \mathrm{m}$
Minimum separation for a microscope, so that images of 2-near by objects are just resolved is given by
$$
\begin{aligned}
d_{\min } & =\frac{1.22 f \lambda}{D}=\frac{1.22 \times 5 \mathrm{~cm} \times 5500 Å}{8 \mathrm{~mm}} \\
& =\frac{1.22 \times 5 \times 10^{-2} \times 5500 \times 10^{-10}}{8 \times 10^{-3}} \\
& =41.9 \times 10^{-7} \mathrm{~m}=4.19 \times 10^{-6} \mathrm{~m}=4.2 \mu \mathrm{m}
\end{aligned}
$$
$$
\begin{aligned}
d_{\min } & =\frac{1.22 f \lambda}{D}=\frac{1.22 \times 5 \mathrm{~cm} \times 5500 Å}{8 \mathrm{~mm}} \\
& =\frac{1.22 \times 5 \times 10^{-2} \times 5500 \times 10^{-10}}{8 \times 10^{-3}} \\
& =41.9 \times 10^{-7} \mathrm{~m}=4.19 \times 10^{-6} \mathrm{~m}=4.2 \mu \mathrm{m}
\end{aligned}
$$
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