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.

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Answer & Solution

Correct Answer

(B) Resolving power will be twice in the oil than it was in the air.

Step-by-step Solution

Detailed explanation

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

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

\((\text { R.P })_{\text {oil }}=(\text { R.P })_{\text {oir }} \times 2\)

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