The numerical aperture of a microscope required to just resolve two points separated by a distance of ${10}^{-4}\text{cm}$ when wavelength of light is $5.8\times {10}^{-5}\text{cm}$ is

The correct option is C $0.35$
The limit of resolution of microscope,

$d=\frac{1.22\lambda }{2\mu \sin \beta }$

Where $\mu \sin \beta$ is called the numerical aperture.

Here, $\lambda =5.8\times {10}^{-5}\text{cm}\text{and}d={10}^{-4}\text{cm}$

$\therefore NA=\frac{1.22\lambda }{2d}=\frac{1.22\times 5.8\times {10}^{-5}}{2\times {10}^{-4}}$

$\therefore NA=0.35$

Hence, $\left(\text{C}\right)$ is the correct answer.