Figure 35-40 shows two isotropic point sources of light (${\mathit{S}}_{\mathbf{1}}$and ${\mathit{S}}_{\mathbf{2}}$) that emit in phase at wavelength 400 nm and at the same amplitude. A detection point P is shown on an x-axis that extends through source ${\mathit{S}}_{\mathbf{1}}$. The phase difference $\mathit{\varphi }$between the light arriving at point P from the two sources is to be measured as P is moved along the x axis from $\mathit{x}\mathbf{=}\mathbf{0}$ out to $\mathit{x}\mathbf{=}\mathbf{+}\mathbf{\infty }$.The results out to ${\mathit{x}}_{\mathbf{s}}\mathbf{=}\mathbf{10}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{7}}\mathbf{}\mathbf{m}$ are given in Fig. 35-41. On the way out to $\mathbf{+}\mathbf{\infty }$ , what is the greatest value of x at which the light arriving at from ${\mathit{S}}_{\mathbf{1}}$is exactly out of phase with the light arriving at P from ${\mathit{S}}_{\mathbf{2}}$?

The phase difference of fringe pattern varies with the path difference. For minimum phase difference path difference should be minimum and vice versa.

The separation is $x_s=10\times {10}^{-7}\text{m}$.

The wavelength of the light is $\lambda =400\mathrm{nm}$$\lambda =400\text{nm}$$\lambda =400\text{nm}$.

The path difference between positions $x=0$ and $x$ is given as:

$\Delta x=\sqrt{d^2+x^2}-x$

Here, d is the separation between sources $S_1$and $S_2$. Its value from the figure 35-40 is $3\lambda$.

The phase difference is given as:

${\varphi }_0-{\varphi }_s=\frac{2\pi }{\lambda }\left(\Delta x\right)$

Here, ${\varphi }_0$and ${\varphi }_s$ are the phase angle for positions $x=0$and $x=x_s$, which are $6\pi$and $5\pi$from the given graph in figure 35-41.

Substitute all the values in equation.

$\begin{array}{rcl}6\pi -5\pi & =& \frac{2\pi }{\lambda }\left(\sqrt{d^2+x^2}-x\right)\\ \sqrt{{\left(3\lambda \right)}^2+x^2}-x& =& \frac{\lambda }{2}\\ \left(9{\lambda }^2+x^2\right)& =& {\left(x+\frac{\lambda }{2}\right)}^2\\ 9{\lambda }^2+x^2& =& \frac{{\lambda }^2}{4}+2\left(\frac{\lambda }{2}\right)x+x^2\end{array}$

$\lambda x=9{\lambda }^2-\frac{{\lambda }^2}{4}$$\lambda x=9{\lambda }^2-\frac{{\lambda }^2}{4}$

$\lambda x=9{\lambda }^2-\frac{{\lambda }^2}{4}$

$\begin{array}{rcl}x& =& \frac{35\lambda }{4}\\ & =& \frac{35\left(400\text{nm}\right)}{4}\\ & =& 3500\text{nm}\end{array}$

Therefore, the maximum value of x for which the light arriving from sources $S_1$ and $S_2$to point out of phase is $3500\text{nm}$.