Wavelength of light used in an optical instrument are $\lambda_{1} = 4000\ \mathring{\mathrm{A}}\ and\ \ \lambda_{2} = 5000\mathring{\mathrm{A}},$ then ratio of their respective resolving powers (corresponding to $\lambda_{1}$ and $\lambda_{2}$) is

Question

Dr. Abubacker · Accepted Answer

(d) Resolving power of an optical instrument is inversely proportional to $\lambda\ i.e.,\ RP \propto \frac{1}{\lambda}\ $
$	herefore\frac{Resolving\ power\ at\ \lambda_{1}}{Resolving\ power\ at\ \lambda_{2}} = \frac{\lambda_{2}}{\lambda_{1}} = \frac{5000}{4000} = 5:4$

Q. Wavelength of light used in an optical instrument are $\lambda_{1} = 4000\ \mathring{\mathrm{A}}\ and\ \ \lambda_{2} = 5000\mathring{\mathrm{A}},$ then ratio of their respective resolving powers (corresponding to $\lambda_{1}$ and $\lambda_{2}$) is

Explanation

Reviewed by Dr. Abubacker

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