Question

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 ${\mathit{S}}_{\mathbf{2}}$?

Short answer

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

Step-by-step solution

Identification of given data

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}$.

Determination of greatest value of x for which the light arriving to point P from both sources is out of phase

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 $3500\text{nm}$.