The resolving power of an optical microscope is evaluated for two distinct wavelengths $\lambda_1 =

Question

The resolving power of an optical microscope is evaluated for two distinct wavelengths $\lambda_1 = 4500 \AA$ and $\lambda_2 = 7500 \AA$. The ratio of the limit of resolution of the microscope for these two wavelengths ($\Delta d_1 : \Delta d_2$) is

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(a) : The limit of resolution of a microscope is given by $\Delta d = \frac{1.22\lambda}{2\mu\sin	heta}$, which implies $\Delta d \propto \lambda$. The ratio of the limits of resolution is directly proportional to the wavelengths: $\frac{\Delta d_1}{\Delta d_2} = \frac{\lambda_1}{\lambda_2} = \frac{4500}{7500} = \frac{3}{5}$.