How to Determine the Liquid's Refractive Index in a Newton's Rings Experiment?

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SUMMARY

The discussion focuses on calculating the refractive index of an unknown liquid using the Newton's rings experiment. Given the diameter of the fourth bright ring in air at 10.0 mm and the diameter in the liquid at 8.45 mm, the refractive index can be determined using the equation xm=√mλfR. The assumption is made that the liquid's refractive index is less than that of glass, which is critical for accurate calculations.

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Homework Statement
Consider a Newton's ring experiment in air. The diameter of the fourth bright rings 10.0mm. When an unknown liquid is poured into the gap between the lens and the support, the diameter of this ring shrinks to 8.45 mm. Calculate the liquid's index of refraction. Assume that the liquid's refractive index is less than the index of refraction of glass (used for the lens and bottom plate).
Relevant Equations
xm=√mλfR.
I have no idea where to start with this. My professor didn't give an example and I am very confused.
 
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ChrisWM said:
Homework Statement:: Consider a Newton's ring experiment in air. The diameter of the fourth bright rings 10.0mm. When an unknown liquid is poured into the gap between the lens and the support, the diameter of this ring shrinks to 8.45 mm. Calculate the liquid's index of refraction. Assume that the liquid's refractive index is less than the index of refraction of glass (used for the lens and bottom plate).
Relevant Equations:: xm=√mλfR.

I have no idea where to start with this. My professor didn't give an example and I am very confused.
What is the equation for the radius of the Nth ring?
https://en.m.wikipedia.org/wiki/Newton's_rings#Theory
 

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