Question

A microscope was initially placed in air (refractive index 1). It is then immersed in oil (refractive index 2). For a light whose wavelength in air is $\lambda$, calculate the change of microscope's resolving power due to oil and choose the correct option

• A. Resolving power will be $\frac{1}{4}$ in the oil than it was in the air
• B. Resolving power will be twice in the oil than it was in the air.
• C. Resolving power will be four times in the oil than it was in the air.
• D. Resolving power will be $\frac{1}{2}$ in the oil than it was in the air.

Answer 1

Answer: (B)

$(R.P{)}_{\text{air }}=\frac{2\sin \theta }{1.22\lambda }$
$(\text{ R.P }{)}_{\text{oil }}=\frac{2\sin \theta }{1.22{\lambda }_{\text{oil }}}=\frac{2\sin \theta \times \mu }{1.22\lambda }$
$(\text{ R.P }{)}_{\text{oil }}=(\text{ R.P }{)}_{\text{air }}\times 2$