JEE Mains · Physics · STD 12 - 10. Wave optics
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 & 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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