A microscope was initially placed in air (refractive index 1 ). It...

Resolving power of microscope in air is $( R . P{)}_{air} = \dfrac{2{\mu}_{air}sin\theta}{1 . 22\lambda} = \dfrac{2sin\theta}{1 . 22\lambda}$

$( R . P{)}_{oil} = \dfrac{2{\mu}_{oil}sin\theta}{1 . 22{\lambda}_{oil}} = \dfrac{2sin\theta\left(2\right)}{1 . 22\lambda}$

${\left(R . P\right)}_{oil} = {\left(R . P\right)}_{air} \times 2$

Thus, resolving power in oil is twice the resolving power in air.