Question

Figure 35-46a shows a lens with radius of curvature lying on a flat glass plate and illuminated from above by light with wavelength l. Figure 35-46b (a photograph taken from above the lens) shows that circular interference fringes (known as Newton’s rings) appear, associated with the variable thickness d of the air film between the lens and the plate. Find the radii r of the interference maxima assuming$\mathit{r}\mathbf{/}\mathit{R}\mathbf{\le }\mathbf{1}$.

Short answer

The radii of the interference maxima is $r=\sqrt{\left(2m+1\right)R\lambda /2}$.

Step-by-step solution

Introduction

Newton's rings is a phenomena in which light reflection between two surfaces—a spherical surface and an adjacent contacting flat surface—creates an interference pattern.

Concept

The interference pattern formed by waves reflected from the upper and lower surface of the air wedge. At the place of condition for the maximum intensity, the thickness of the wedge is .

The following figure shows the experimental setup of Newton’s ring.

Expression for the condition of constructive interference is,

$2d=\left(m+\frac{1}{2}\right)\lambda$

Here, d is the thickness of the wedge, m is the order of the fringe pattern, and $\lambda$ is the wavelength of light in air.

Rearrange the above expression for d.

$d=\left(2m+1\right)\frac{\lambda }{4}$ ...(1)

Here, d thickness of the wedge.

Express the relation form the diagram.

$DC\times ED=BD\times DA$

Here, $DC,ED,BD$ and DA are lengths.

Substitute r for both DC and ED, d for BD, and R for DA find d.

$\begin{array}{l}r\times r=d\times R\\ R=\frac{r^2}{d}\end{array}$$\begin{array}{rcl}r\times r& =& d\times R\\ R& =& \frac{r^2}{d}\end{array}$ ...(2)

Here r is the radius of the Newton’s ring,

And R is the radius of curvature of the lens.

Solution

We use condition $r/R\le 1$ to simplify the expression in equation since that

$2R-d\approx 2R$

Rearrange the expression in equation (2) for d.

$d=\frac{r^2}{2R}$ ...(3)

Here thickness of the wedge.

Solve equation (1) and (3) for d.

$\begin{array}{rcl}\frac{r^2}{2R}& =& \frac{\left(2m+1\right)\lambda }{4}\\ r^2& =& \frac{\left(2m+1\right)}{2}\lambda R\\ r& =& \sqrt{\frac{\left(2m+1\right)R\lambda }{2}}\text{for m = 0,1,2,3,}.......\end{array}$

Therefore, the radii of the interference maxima is $\boxed{r=\sqrt{\left(2m+1\right)R\lambda /2}}$.