JEE Main · 2022 · 27 Jul 2022 · Shift 1 · Geometrical Optics

The resolving power of a microscope is proportional to its numerical aperture, which carries the refractive index of the medium between the object and the objective: $R.P.=\frac{2\mu \sin \theta }{1.22\lambda }$, where $\lambda$ is the wavelength measured in air. Immersing the instrument in oil changes nothing in this expression except the index $\mu$, which rises from $1$ to $2$. Because the resolving power depends on $\mu$ to the first power, doubling the refractive index of the medium simply doubles the resolving power, so in the oil it becomes twice what it was in air.