In a Newton ring’s experiment the diameter of 5th dark ring is 0.336 cm and the diameter of 15th dark ring is 0.590 cm. Find the radius of curvature of a plano convex lens, if the wavelength of light used is 5890 $\unicode{x212B}$

It is known that due to very small radii the first few rings of the Newton's rings pattern are not distinguishable. It is not possible to identify the exact order (n) of any ring.

Hence, an arbitrary ring, say nth dark ring is studied with its radius taken as $r_n$ and diameter measured as $D_n$ with the help of a travelling microscope.

The nth dark ring is represented by

$$\begin{aligned} r_n^2 &= R_n \ \lambda \\ \text{and } D_n^2 &= 4 R_n \ \lambda \text{ ----- (1)} \end{aligned}$$

The diameter of a larger arbitrary dark ring, say, the (n+m)th ring is simultaneously measured as

$$D_{n+m}^2 = 4 R(n+m) \lambda \text{ ----- (2)}$$

The difference of the two equations become

$$\begin{aligned} D_{n+m}^2 - D_n^2 &= 4 R m \lambda \\ \therefore R &= \frac{D_{n+m}^2 - D_n^2}{4 m \lambda} \end{aligned}$$

So, if the wavelength of the light is known, then the radius of curvature of the plano convex lens can be calculated.

Numerical:

$D_5 = 0.336 cm$

$D_15 = 0.590 cm$

$\lambda = 5890 A^o = 5890 \times 10^{-10} m$

$\mu = 1$

$\begin{aligned} R &= \frac{D_{n+m}^2 - D_0^2 }{4m\lambda} \\ &= \frac{D_{15}^2 - D_5^2}{4 \times 10 \times \lambda} \\ &= \frac{0.590^2 - 0.336^2}{4 \times 10 \times 5890 \times 10^{-8}} \\ R &= 99.88 cm \end{aligned}$