91Ó°ÊÓ

The distance between the sources, $d=6\mathrm{λ}$.

The distance from $S_2$ to screen, $D=20\mathrm{λ}$.

The expression to calculate the number of the wavelength in the path is given as follows.

$\mathit{N}\mathbf{=}\frac{\mathbf{L}}{\mathbf{λ}}$ ….. (1)

Here, is the source and point and is the wavelength.

The expression to calculate the phase difference is given as follows.

$\mathit{ϕ}\mathbf{=}{\mathit{N}}_{\mathbf{1}}\mathbf{-}{\mathit{N}}_{\mathbf{2}}$ …… (2)

Here,${\mathit{N}}_{\mathbf{1}}$ is the number of wavelengths in first path and ${\mathit{N}}_{\mathbf{2}}$is the number of wavelengths in second path.

Consider the figure shown below.

Calculate the distance between the first source and point P as,

$$\begin{gathered} L_1^2={\left(d+D\right)}^2+y_p^2 \\ {L}_1=\sqrt{{\left(d+D\right)}^2+y_p^2} \end{gathered}$$

Calculate the number of the wavelength in the first path,

$N_1=\frac{L_1}{\mathrm{λ}}$

Substitute $\sqrt{{\left(d+D\right)}^2+y_p^2}$for $L_1$ into above equation.

$N_1=\frac{\sqrt{{\left(d+D\right)}^2+y_p^2}}{\mathrm{λ}}$

Calculate the distance between the second source and point P as,

$$\begin{gathered} L_2^2=D^2+y_p^2 \\ {L}_2=\sqrt{D^2+y_p^2} \end{gathered}$$

Calculate the number of the wavelength in the second path,

$N_2=\frac{L_2}{\mathrm{λ}}$

Substitute $\sqrt{D^2+y_p^2}$ for $L_2$ into above equation.

$N_2=\frac{\sqrt{D^2+y_p^2}}{\mathrm{λ}}$

Calculate the phase difference.

Substitute $\frac{\sqrt{{\left(d+D\right)}^2+y_p^2}}{\mathrm{λ}}$ for $N_1$ and $\frac{\sqrt{D^2+y_p^2}}{\mathrm{λ}}$ for $N_2$ into equation (1).

$$\begin{gathered} \mathrm{Φ}=\frac{\sqrt{{\left(d+D\right)}^2+y_p^2}}{\mathrm{λ}}-\frac{\sqrt{D^2+y_p^2}}{\mathrm{λ}} \\ \mathrm{Φ}=\frac{\sqrt{{\left(d+D\right)}^2+y_p^2}-\sqrt{D^2+y_p^2}}{\mathrm{λ}}......\left(2\right) \end{gathered}$$

The phase difference is minimum at $y_p=∞$.

$\begin{array}{rcl}\mathrm{Φ}& =& \frac{\sqrt{{\left(d+D\right)}^2+{\left(∞\right)}^2}-\sqrt{D^2+{\left(∞\right)}^2}}{\mathrm{λ}}\\ & =& 0\end{array}$

Hence, the value of $y_p$at which phase difference is minimum is $∞$.

The phase difference is directly proportional to the number of wavelengths in the path and the phase difference is minimum at $y_p$ which results the zero-phase difference.

The phase difference is multiple $1/\mathrm{λ}$. Therefore, the multiple of the $\mathrm{λ}$ is zero.

Hence the multiple of the $\mathrm{λ}$ that gives the minimum phase difference is zero.

Recall the equation (2).

$\mathrm{Φ}=\frac{\sqrt{{\left(d+D\right)}^2+y_p^2}-\sqrt{D^2+y_p^2}}{\mathrm{λ}}$

Differentiate the above equation with respect to $y_p$.

$\begin{array}{rcl}\frac{d\mathrm{ϕ}}{d{y}_p}& =& \frac{2y_p{\left[{\left(d+D\right)}^2+{y^2}_p\right]}^{-1/2}-2y_p{\left(D^2+{y^2}_p\right)}^{-1/2}}{\mathrm{λ}}\\ & =& \frac{2y_p\left\{{\left[{\left(d+D\right)}^2+{y^2}_p\right]}^{-1/2}-{\left(D^2+{y^2}_p\right)}^{-1/2}\right\}}{\mathrm{λ}}\end{array}$

Substitute 0 for $\frac{d\mathrm{Φ}}{d{y}_p}$ into above equation.

$\begin{array}{rcl}\frac{2y_p\left\{{\left[{\left(d+D\right)}^2+{y^2}_p\right]}^{-1/2}-{\left(D^2+{y^2}_p\right)}^{-1/2}\right\}}{\mathrm{λ}}& =& 0\\ 2y_p\left\{{\left[{\left(d+D\right)}^2+{y^2}_p\right]}^{-1/2}-{\left(D^2+{y^2}_p\right)}^{-1/2}\right\}& =& 0\end{array}$

From the above equation,

$\begin{array}{rcl}2y_p& =& 0\\ y_p& =& 0\end{array}$

Hence the maximum possible phase difference occurs at $y_p=0$.

Calculate the maximum phase difference.

Substitute 0 for $y_p$ into equation (2).

$\begin{array}{rcl}\mathrm{Φ}& =& \frac{\sqrt{{\left(d+D\right)}^2+0^2}-\sqrt{D^2+0^2}}{\mathrm{λ}}\\ & =& \frac{\sqrt{{\left(d+D\right)}^2}-\sqrt{D^2}}{\mathrm{λ}}\\ & =& \frac{d+D-D}{\mathrm{λ}}\\ & =& \frac{d}{\mathrm{λ}}\end{array}$

Substitute $6\mathrm{λ}$ for d into above equation.

$\begin{array}{rcl}\mathrm{Φ}& =& \frac{6\mathrm{λ}}{\mathrm{λ}}\\ & =& 6\end{array}$

Hence, the maximum phase difference is 6.

Calculate the phase difference at $y_p=d$.

Substitute d for $y_p$ into equation (2).

$\mathrm{Φ}=\frac{\sqrt{{\left(d+D\right)}^2+d^2}-\sqrt{D^2+d^2}}{\mathrm{λ}}$

Substitute $6\mathrm{λ}$ for d and $20\mathrm{λ}$ for D into above equation.

$\begin{array}{rcl}\mathrm{Φ}& =& \frac{\sqrt{{\left(6\mathrm{λ}+20\mathrm{λ}\right)}^2+{\left(6\mathrm{λ}\right)}^2}-\sqrt{{\left(20\mathrm{λ}\right)}^2+{\left(6\mathrm{λ}\right)}^2}}{\mathrm{λ}}\\ & =& \frac{\sqrt{{\left(26\mathrm{λ}\right)}^2+{\left(6\mathrm{λ}\right)}^2}-\sqrt{400{\mathrm{λ}}^2+36{\mathrm{λ}}^2}}{\mathrm{λ}}\\ & =& \frac{\sqrt{676{\mathrm{λ}}^2+36{\mathrm{λ}}^2}-\sqrt{400{\mathrm{λ}}^2+36{\mathrm{λ}}^2}}{\mathrm{λ}}\\ & =& \frac{\sqrt{676{\mathrm{λ}}^2+36{\mathrm{λ}}^2}-\sqrt{400{\mathrm{λ}}^2+36{\mathrm{λ}}^2}}{\mathrm{λ}}\end{array}$

Solve further as,

$\begin{array}{rcl}\mathrm{Φ}& =& \frac{\sqrt{712{\mathrm{λ}}^2}-\sqrt{436{\mathrm{λ}}^2}}{\mathrm{λ}}\\ & =& \frac{\mathrm{λ}\left(\sqrt{712}-\sqrt{436}\right)}{\mathrm{λ}}\\ & =& \sqrt{712}-\sqrt{436}\\ & =& 5.8\end{array}$

Hence, the phase difference when $y_p=d$ is 5.8.

The maximum phase difference is calculated in the part (d) is 6 and phase difference when $y_p=d$ is calculated in part (e) is 5.8 and it is closer to the maximum value than the half integer.

Hence, the intensity when $y_p=d$ is intermediate but closer to maximum.