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(C) The minimum separation $(d_{\min})$ between two objects that can be just resolved by a microscope is given by the formula:$d_{\min} = \frac{1.22 f

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$4.2$

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(C) The minimum separation $(d_{\min})$ between two objects that can be just resolved by a microscope is given by the formula:$d_{\min} = \frac{1.22 f \lambda}{D}$Given:$f = 5 	ext{ cm} = 5 	imes 10^{-2} 	ext{ m}$$\lambda = 5500 	ext{ Å} = 5500 	imes 10^{-10} 	ext{ m}$$D = 8 	ext{ mm} = 8 	imes 10^{-3} 	ext{ m}$Substituting the values:$d_{\min} = \frac{1.22 	imes 5 	imes 10^{-2} 	imes 5500 	imes 10^{-10}}{8 	imes 10^{-3}}$$d_{\min} = \frac{1.22 	imes 5 	imes 5500 	imes 10^{-12}}{8 	imes 10^{-3}}$$d_{\min} = \frac{33550 	imes 10^{-12}}{8 	imes 10^{-3}}$$d_{\min} = 4193.75 	imes 10^{-9} 	ext{ m} \approx 4.19 	imes 10^{-6} 	ext{ m}$$d_{\min} \approx 4.2 \mu	ext{m}$

$A$ microscope has an objective of aperture $8 \text{ mm}$ and focal length of $5 \text{ cm}$. The minimum separation between two objects to be just resolved by the microscope is (wavelength of light used $= 5500 \text{ Å}$) (in $\mu\text{m}$)

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