In Fig. 35-39, two isotropic point sources S₁ and S₂ emit light in phase at wavelength $\mathit{\lambda }$ and at the same amplitude. The sources are separated by distance $\mathbf{2}\mathit{d}\mathbf{=}\mathbf{6}\mathit{\lambda }$. They lie on an axis that is parallel to an x axis, which runs along a viewing screen at distance $\mathit{D}\mathbf{=}\mathbf{20}\mathbf{.}\mathbf{0}\mathit{\lambda }$. The origin lies on the perpendicular bisector between the sources. The figure shows two rays reaching point P on the screen, at position${\mathit{x}}_{\mathbf{P}}$. (a) At what value of ${\mathit{x}}_{\mathbf{P}}$do the rays have the minimum possible phase difference? (b) What multiple of$\mathit{\lambda }$ gives that minimum phase difference? (c) At what value of${\mathit{x}}_{\mathbf{P}}$do the rays have the maximum possible phase difference? What multiple of $\mathit{\lambda }$ gives (d) that maximum phase difference and (e) the phase difference when ${\mathit{x}}_{\mathbf{P}}\mathbf{=}\mathbf{6}\mathit{\lambda }$ ? (f) When ${\mathit{x}}_{\mathbf{P}}\mathbf{=}\mathbf{6}\mathit{\lambda }$, is the resulting intensity at point P maximum, minimum, intermediate but closer to maximum, or intermediate but closer to minimum?

The separation between both sources is $2d=6\lambda$

The path difference for maximum phase difference is $x_P=6\lambda$

The distance of both sources from screen is $D=20\lambda$

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

The path difference for minimum possible phase difference is given as:

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

Here, m is the order of fringe for minimum intensities at various positions and its minimum value for minimum possible path difference is zero.

Substitute all the values in equation.

$\begin{array}{rcl}{x}_P& =& \left(0+\frac{1}{2}\right)\lambda \\ x_P& =& \frac{\lambda }{2}\end{array}$

Therefore, the path difference for minimum possible phase difference is $\frac{\lambda }{2}$.

The path difference for minimum possible phase difference is given as:

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

Here all the order of fringes from 0, 1, 2 ….. gives the $\frac{1}{2}$, $\frac{3}{2}$, $\frac{5}{2}$……… This means all the multiples for minimum phase difference are odd multiples of wavelength.

Therefore, the odd multiple of wavelength provides the minimum phase difference.

Maximum value of order of fringe for maximum path difference is given as:

$d\sin \theta =m\lambda$

Here,m is the order of fringe for maximum intensities at various positions and the maximum value will corresponding to maximum value of $\sin \theta$ which is 1.

Substitute all the values in equation.

$\begin{array}{rcl}\left(3\lambda \right)\left(1\right)& =& m\lambda \\ m& =& 3\end{array}$

The path difference for maximum possible phase difference is given as:

$x_P=m\lambda$

Substitute all the values in equation.

$\begin{array}{rcl}{x}_P& =& \left(3\right)\lambda \\ x_P& =& 3\lambda \end{array}$

Therefore, the path difference for maximum possible phase difference is $3\lambda$.

The maximum possible phase difference is given as:

${\varphi }_P=\left(\frac{2\pi }{\lambda }\right)x_P$

Substitute all the values in the above equation.

$\begin{array}{rcl}{\varphi }_P& =& \left(\frac{2\pi }{\lambda }\right)\left(3\lambda \right)\\ {\varphi }_P& =& 6\pi \end{array}$

Therefore, the maximum possible phase difference is $6\pi$.

The maximum possible phase difference is given as:

${\varphi }_P=\left(\frac{2\pi }{\lambda }\right)x_P$

Substitute all the values in the above equation.

$\begin{array}{rcl}{\varphi }_P& =& \left(\frac{2\pi }{\lambda }\right)\left(6\lambda \right)\\ {\varphi }_P& =& 12\pi \end{array}$

Therefore, the maximum possible phase difference is $12\pi$.

The resulting intensity at point P for $x_P=6\lambda$ will be maximum because the even multiples of path difference produce bright fringes.

Therefore, the intensity at point P is maximum.