Question

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?

Short answer

(a) The path difference for minimum possible phase difference is $\frac{\lambda }{2}$.

(b) The odd multiple of wavelength provides the minimum phase difference.

(c) The path difference for maximum possible phase difference is $3\lambda$.

(d) The maximum possible phase difference is $6\pi$.

(e) The maximum possible phase difference is $12\pi$.

(f) The intensity at point P is maximum.

Step-by-step solution

Identification of given data

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.

Step 2(a): Determination of path difference for minimum possible phase difference

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 $\frac{\lambda }{2}$.

Step 3(b): Determination of multiple of wavelength for minimum phase difference

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.

Step 4(c): Determination of path difference for maximum possible 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 $3\lambda$.

Step 5(d): Determination of maximum phase difference

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 $6\pi$.

Step 6(e): Determination of maximum phase difference

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 $12\pi$.

Step 7(f): Determination of maximum phase difference

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.