In a Newton’s ring arrangement, a drop of liquid of refractive index 1.33 is placed in between the lens and plate. The diameter of 9th dark ring is found to be 0.58 cm. Obtain the radius of curvature, if the wavelength of light is 6000 A°.

Given: \(\mu=1.33\), \(n=9\), \(D_9=0.58\) cm \(=5.8\times10^{-3}\) m, \(\lambda=6000\) Å \(=6\times10^{-7}\) m.

Formula: For dark rings in reflected light with a liquid film, \(2\mu t=n\lambda\) and \(t=r^2/2R\), which gives

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

Working:

\[ R=\frac{1.33\times(5.8\times10^{-3})^2}{4\times9\times6\times10^{-7}}=\frac{1.33\times3.364\times10^{-5}}{2.16\times10^{-5}}=2.07\ \text{m} \]

Radius of curvature \(R\approx2.07\) m (≈ 207 cm)

Check: \(4\times9\times6\times10^{-7}\times2.0714/1.33=3.364\times10^{-5}\) m² \(=D_9^2\). The answer checks out.