91Ó°ÊÓ

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 $x_s=10×{10}^{-7}\text{m}$.

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

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

$\mathrm{Δ}x=\sqrt{d^2+x^2}-x$

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

The phase difference is given as:

${\mathrm{ϕ}}_0-{\mathrm{ϕ}}_s=\frac{2\mathrm{π}}{\mathrm{λ}}\left(\mathrm{Δ}x\right)$

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

Substitute all the values in equation.

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

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

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

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

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