Resolving Power question from JEE Main 2014 (19 Apr, Shift 1)

Answer: (c)

1. Resolving power of a microscope $= \dfrac{2n\sin\theta}{1.22\lambda}$, where $\theta$ is set by the cone of light the objective accepts.
2. So it increases with $n$ and with the angle term $\sin(\text{angle})$; it falls if $n$ or $\beta$ decreases.
3. Of the options, only $n\sin 2\beta$ increases with both $n$ and $\beta$ (for small angles).
4. So the answer is (c) increases with increasing value of $n\sin 2\beta$.