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We have one experimental formula connecting the wavelength of radiation in air as

Lambdaair =( r2n+m - r2n )/ m R

Now we have pour the liquid in between the lens and glass plate. So intead air film we have a liquid film. So wavelength Lambdaliq can be found by the similar expression

Lambdaliq =( s2n+m - s2n )/ m R

Refractive index = lambda in air / lambda in liq

Hence refractive index = ( r2n+m - r2n )/ (s2n+m - s2n )

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Q: How do you find the wavelength and refractive index of a liquid used by using newton's ring.?
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Related questions

What color of light has the highest refractive index?

The refractive index is inversely proportional to the wavelength, so the shorter the wavelength (the higher the frequency, or the more "blue" the light) the higher the refractive index. Conversely, the longer the wavelength (the lower the frequency, or the more "red" the light), the lower the refractive index. Therefore as wavelength of blue in less the refractive index will be maximum. For more information, follow the related link below.


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The refractive index depends on the wavelength of the radiation. Traditionally the sodium line NaD20 (589,3 nm) is used for measurements (20 is the temperature on the Celsius scale).


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