The lens in a Newton’s rings experiment (see problem 75) has diameter 20 mm and radius of curvature $\mathit{R}\mathbf{=}\mathbf{5}\mathbf{.}\mathbf{0}\mathbf{}\mathit{m}$. For $\mathit{A}\mathbf{=}\mathbf{589}\mathbf{}\mathbf{nm}$ in air, how many bright rings are produced with the setup (a) in air and
(b) immersed in water $\mathbf{(}\mathit{n}\mathbf{=}\mathbf{1}\mathbf{.}\mathbf{33}\mathbf{)}$?

Newton's rings arise from the interference of light. The phenomenon of interference of light waves is obtained from monochromatic and coherent rays i.e. rays of same frequency and constant phase difference.

(a)

In Newton’s ring experiment, as air film of variable thickness is obtained between glass plate and convex surface .in this case, the radii of fringes for interference maxima is calculated as follows,

$\begin{array}{rcl}r& =& \sqrt{\left(m+\frac{1}{2}\right)\lambda R}\\ \frac{r^2}{\lambda R}& =& m+\frac{1}{2}\\ m& =& \frac{r^2}{\lambda R}-\frac{1}{2}\end{array}$

Here, m represents number of fringes, r is radius of fringe, R is radius of curvature and $\lambda$is wavelength of incident light.

Substitute 10 mm for r, 589 nm for $\lambda$and 5 m for R.

$\begin{array}{rcl}m& =& \frac{{\left(10\text{mm}\times \frac{{\text{10}}^{-3}\text{m}}{1\text{mm}}\right)}^2}{\left(5\text{m}\right)\left(589\text{mm}\times \frac{{10}^{-9}\text{m}}{1\text{nm}}\right)}-\frac{1}{2}\\ & =& 33\end{array}$

Therefore, the number of bright rings produced is 33.

(b)

Consider the following formula,

$m=\frac{r^2}{{\lambda }_{n}R}-\frac{1}{2}$

If the arrangement were immersed in water $\left(n_w=1.33\right)$ then the wavelength is changed.

The new wavelength is calculated as follows,

${\lambda }_n=\frac{\lambda }{n_w}$

Substitute $\frac{\lambda }{n_w}$ for ${\lambda }_n$.

$$\begin{gathered} m=\frac{r^2}{\left(\frac{\lambda }{n_w}\right)R}-\frac{1}{2} \\ m=\frac{r^2{n}_w}{\lambda R}-\frac{1}{2} \end{gathered}$$

Substitute 10 mm for r, 589 nm for $\lambda$, 5 m for R and 1.33 for $n_w$.

$\begin{array}{rcl}m& =& \frac{{\left(10\text{mm}\times \frac{{\text{10}}^{-3}\text{m}}{1\text{mm}}\right)}^2\left(1.33\right)}{\left(5\text{m}\right)\left(589\text{mm}\times \frac{{10}^{-9}\text{mm}}{1\text{nm}}\right)}-\frac{1}{2}\\ & =& 45\end{array}$

Therefore, the number of bright rings produced is 45.