Wavelengths of light used in an optical instr | vedclass

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(D) The resolving power $(RP)$ of an optical instrument is given by the formula:$RP = \frac{2 \mu \sin 	heta}{1.22 \lambda}$From this expression,we c

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$5: 4$

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(D) The resolving power $(RP)$ of an optical instrument is given by the formula:$RP = \frac{2 \mu \sin 	heta}{1.22 \lambda}$From this expression,we can see that the resolving power is inversely proportional to the wavelength of light used:$RP \propto \frac{1}{\lambda}$Therefore,the ratio of the resolving powers ($R_1$ and $R_2$) corresponding to wavelengths $\lambda_1$ and $\lambda_2$ is:$\frac{R_1}{R_2} = \frac{\lambda_2}{\lambda_1}$Given $\lambda_1 = 4000 \ Å$ and $\lambda_2 = 5000 \ Å$,we substitute these values:$\frac{R_1}{R_2} = \frac{5000}{4000} = \frac{5}{4}$Thus,the ratio is $5: 4$.

Wavelengths of light used in an optical instrument are $\lambda_1 = 4000 \ Å$ and $\lambda_2 = 5000 \ Å$. The ratio of their respective resolving powers (corresponding to $\lambda_1$ and $\lambda_2$) is:

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