For the same objective,find the ratio of the least separation between two points to be distinguished by a microscope for light of $5000\,\mathring{A}$ and electrons accelerated through $100\,V$ used as the illuminating substance.

(D) The limit of resolution $d$ for a microscope is given by $d = \frac{1.22 \lambda}{2 n \sin \beta}$.
Since the objective is the same,$n$ and $\beta$ are constant,so $d \propto \lambda$.
Therefore,the ratio of the least separation is $\frac{d_1}{d_2} = \frac{\lambda_1}{\lambda_2}$.
For light,$\lambda_1 = 5000 \times 10^{-10} \, m = 5 \times 10^{-7} \, m$.
For electrons accelerated through $V = 100 \, V$,the de Broglie wavelength is $\lambda_2 = \frac{h}{\sqrt{2meV}}$.
Using $\lambda_2 \approx \frac{12.27}{\sqrt{V}} \, \mathring{A} = \frac{12.27}{\sqrt{100}} \, \mathring{A} = 1.227 \, \mathring{A} = 1.227 \times 10^{-10} \, m$.
Now,the ratio $\frac{d_1}{d_2} = \frac{5000 \times 10^{-10}}{1.227 \times 10^{-10}} \approx 4075$.