Question

In Fig. 35-37, two isotropic point sources S1 and S2 emit identical light waves in phase at wavelength$\mathit{\lambda }$. The sources lie at separation on an x axis, and a light detector is moved in a circle of large radius around the midpoint between them. It detects 30points of zero intensity, including two on the xaxis, one of them to the left of the sources and the other to the right of the sources. What is the value of $\raisebox{1ex}{$\mathbf{d}$}\!\left/ \!\raisebox{-1ex}{$\mathbf{\lambda }$}\right.$?

Short answer

The value $\raisebox{1ex}{$d$}\!\left/ \!\raisebox{-1ex}{$\lambda $}\right.$ is $m+\frac{1}{2}$ where $m=0,1,2,3...$

Step-by-step solution

Given data

The wavelength the waves emitted from the sources is$\lambda$ .

The separation between the two sources is d.

Path difference for destructive interference

The path difference for destructive interference of two waves having wavelength $\lambda$$∆L=\left(m+\frac{1}{2}\right)\lambda m=0,1,2,3...$ ..... (1)

Determining the ratio dλ

At any point on the x axis the path difference between the two waves is

$\begin{array}{rcl}\Delta L& =& x+d-x\\ & =& d\end{array}$

From equation (1), for destructive interference

$\begin{array}{rcl}d& =& \left(m+\frac{1}{2}\right)\lambda \\ \frac{d}{\lambda }& =& m+\frac{1}{2}\end{array}$

Hence, the ratio is $m+\frac{1}{2}$ where $m=0,1,2,3...$.