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  • 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 <img alt="\mathrm{\lambda,}" src="https://en

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 \mathrm{\lambda,} calculate the change of microscope's resolving power due to oil and choose the correct option.

Option: 1

Resolving power will be \frac{1}{4}  in the oil than it was in the air. 


Option: 2

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


Option: 3

Resolving power will be four times in the oil than it was in the air.


Option: 4

Resolving power will be \frac{1}{2} in the oil than it was in the air. 


Answers (1)

best_answer

\mathrm{\mu_{\text {air }}=1 }.

\mathrm{\mu_{\text {oil }}=2 }

\mathrm{\lambda_{\text {air }}=\lambda}
Resolving power of a microscope is
\mathrm{R \cdot P \cdot =\frac{2 \mu \sin \theta}{\lambda} }

\mathrm{\frac{(R \cdot P)_{\text {air }}}{(R \cdot P)_{\text {oil }}} =\frac{\mu_{\text {air }}}{\lambda_{\text {air }}} \times \frac{\lambda_{\text {oil }}}{\mu_{\text {oil }}} }

\mathrm{\frac{(R \cdot P \cdot)_{\text {air }}}{(R \cdot P)_{\text {oil }}} =\left(\frac{\mu_{\text {ais }}}{\mu_{\text {oil }}}\right)^2 }

\mathrm{(R \cdot P)_{\text {oil }} =4(R \cdot P)_{\text {air }}}
Hence 3 is correct option



 

Posted by

Divya Prakash Singh

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