In a Newton's rings experiment the diameter of the 10th dark ring is 0.50 cm with air between the lens and the plate. When a transparent liquid is introduced between them, the diameter of the same ring becomes 0.44 cm. Find the refractive index of th…

Answer: \(\mu\approx1.29\); \(R\approx1.06\) m.

For the \(n^{th}\) dark ring in reflected light with a film of refractive index \(\mu\):

\[D_n^2=\frac{4n\lambda R}{\mu}\]

Refractive index. For the same ring (\(n=10\)) with air (\(\mu=1\)) and with liquid:

\[\frac{D_{air}^2}{D_{liq}^2}=\frac{4n\lambda R}{4n\lambda R/\mu}=\mu\ \Rightarrow\ \mu=\left(\frac{0.50}{0.44}\right)^2=(1.1364)^2=1.29\]

Radius of curvature (from the air reading): \(D_{10}=0.50\) cm \(=5\times10^{-3}\) m.

\[R=\frac{D_{10}^2}{4n\lambda}=\frac{(5\times10^{-3})^2}{4\times10\times5890\times10^{-10}}=\frac{2.5\times10^{-5}}{2.356\times10^{-5}}=1.06\ \text{m}\]

The rings shrink when a liquid is introduced because the optical path in the film increases by the factor \(\mu\).