Two points separated by a distance of $0.1\text{mm}$ can just be inspected in a microscope when light of wavelength $6000\dot{A}$ is used. If the light of wavelength $4800\dot{A}$ is used, then the limit of resolution will become

The correct option is D $0.08\text{mm}$
The limit of resolution of the microscope is given by,

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

Where, $\mu \to \text{refractive index of the medium between object and objective.}$
$\beta \to \text{angle subtended by the radius of the objective with object}$

In the given situation, $\mu$ and $\sin \beta$ are constant,

$\Rightarrow d_{min}\propto \lambda$

$\Rightarrow \frac{d_2}{d_1}=\frac{{\lambda }_2}{{\lambda }_1}$

$\therefore$ The new limit of resolution

$d_2=\frac{{\lambda }_2}{{\lambda }_1}\times d_1$

$d_2=\frac{4800}{6000}\times 0.1\text{mm}=0.08\text{mm}$

Hence, option $\left(\text{D}\right)$ is correct.