In Newton’s rings experiment the diameter of 5th ring was 0.336 cm and the diameter of 15th ring was 0.590 cm. Find the radius of curvature of plano convex lens if the wavelength of light used is 5890 Å.

Given: \(D_5=0.336\) cm, \(D_{15}=0.590\) cm, so p = 10. \(\lambda=5890\) Å \(=5.89\times10^{-7}\) m. Air film (μ = 1).

Formula: \(D_{n+p}^2-D_n^2=4p\lambda R\Rightarrow R=\dfrac{D_{n+p}^2-D_n^2}{4p\lambda}\)

Working:

\(D_{15}^2=0.3481\) cm² and \(D_5^2=0.112896\) cm²

Difference \(=0.235204\) cm² \(=2.35204\times10^{-5}\) m²

\[ R=\frac{2.35204\times10^{-5}}{4\times10\times5.89\times10^{-7}}=\frac{2.35204\times10^{-5}}{2.356\times10^{-5}}=0.998\ \text{m} \]

Radius of curvature \(R\approx0.998\) m ≈ 1 m (99.8 cm)

Check: \(40\times5.89\times10^{-7}\times0.9983=2.352\times10^{-5}\) m². The answer checks out.