The value of numerical aperture of the object | vedclass

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(C) The resolving power of a microscope is defined by the minimum separation $d$ between two points that can be resolved.The formula for the minimum s

Vedclass · Accepted Answer

$0.24$

Vedclass · Answer

(C) The resolving power of a microscope is defined by the minimum separation $d$ between two points that can be resolved.The formula for the minimum separation $d$ is given by:$d = \frac{0.61 \lambda}{	ext{NA}}$Given:$\lambda = 5000\,\mathring{A} = 5000 	imes 10^{-10}\,	ext{m} = 5 	imes 10^{-7}\,	ext{m}$$	ext{NA} = 1.25$Substituting the values:$d = \frac{0.61 	imes 5 	imes 10^{-7}}{1.25}$$d = \frac{3.05 	imes 10^{-7}}{1.25}$$d = 2.44 	imes 10^{-7}\,	ext{m}$Converting to micrometers $(\mu m)$:$d = 2.44 	imes 10^{-7} 	imes 10^6\,\mu m = 0.244\,\mu m \approx 0.24\,\mu m$Thus, the correct option is $C$.

The value of numerical aperture of the objective lens of a microscope is $1.25$. If light of wavelength $5000\,\mathring{A}$ is used, the minimum separation between two points, to be seen as distinct, will be....$\mu m$

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