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$