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(B) The resolving power $(RP)$ of an optical microscope is given by the formula:$RP = \frac{2\mu \sin 	heta}{\lambda}$where $\mu$ is the refractive i

Vedclass · Accepted Answer

$3:2$

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(B) The resolving power $(RP)$ of an optical microscope is given by the formula:$RP = \frac{2\mu \sin 	heta}{\lambda}$where $\mu$ is the refractive index of the medium and $	heta$ is the semi-vertical angle of the cone of light.From the formula,it is clear that $RP \propto \frac{1}{\lambda}$.For wavelength $\lambda_1 = 4000\,\mathring{A}$,the resolving power is $RP_1 = \frac{k}{4000}$ (where $k = 2\mu \sin 	heta$).For wavelength $\lambda_2 = 6000\,\mathring{A}$,the resolving power is $RP_2 = \frac{k}{6000}$.The ratio of the resolving powers is:$\frac{RP_1}{RP_2} = \frac{k/4000}{k/6000} = \frac{6000}{4000} = \frac{3}{2}$.Thus,the ratio is $3:2$.

The ratio of resolving powers of an optical microscope for two wavelengths $\lambda_1 = 4000\,\mathring{A}$ and $\lambda_2 = 6000\,\mathring{A}$ is

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