Question

In Figure 35-50, two isotropic point sources ${\mathit{S}}_{\mathbf{1}}$and ${\mathit{S}}_{\mathbf{2}}$emit light in phase at wavelength $\mathit{\lambda }$and at the same amplitude. The sources are separated by distance $\mathit{d}\mathbf{=}\mathbf{6}\mathbf{.}\mathbf{00}\mathit{\lambda }$on an x axis. A viewing screen is at distance $\mathit{D}\mathbf{=}\mathbf{20}\mathbf{.}\mathbf{0}\mathit{\lambda }$from ${\mathit{S}}_{\mathbf{2}}$and parallel to the y axis. The figure shows two rays reaching point P on the screen, at height ${\mathit{y}}_{\mathbf{p}}$. (a) At what value of 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 ${\mathit{y}}_{\mathbf{p}}$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 ${\mathit{y}}_{\mathbf{p}}\mathbf{=}\mathit{d}$? (f) When ${\mathit{y}}_{\mathbf{p}}\mathbf{=}\mathit{d}$, is the resulting intensity at point P maximum, minimum, intermediate but closer to maximum, or intermediate but closer to minimum?

Short answer

(a) The value of $y_p$ at which phase difference is minimum is $\infty$.

(b) The multiple of the $\lambda$that gives the minimum phase difference is zero.

(c) The maximum possible phase difference occurs at $y_p=0$.

(d) The maximum phase difference is 6.

(e) The phase difference when $y_p=d$ is 5.8.

(f) The intensity when $y_p=d$ is intermediate but closer to maximum.

Step-by-step solution

Write the given data from the question

The distance between the sources, $d=6\lambda$.

The distance from $S_2$ to screen, $D=20\lambda$.

Determine the formulas to calculate the value of  yp, multiple of  λ, and phase difference

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

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

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

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

$\mathit{\varphi }\mathbf{=}{\mathit{N}}_{\mathbf{1}}\mathbf{-}{\mathit{N}}_{\mathbf{2}}$ …… (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.

(a) Calculate the value of  yp for the minimum possible phase difference

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,

$N_1=\frac{L_1}{\lambda }$

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

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

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,

$N_2=\frac{L_2}{\lambda }$

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

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

Calculate the phase difference.

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

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

The phase difference is minimum at $y_p=\infty$.

$\begin{array}{rcl}\Phi & =& \frac{\sqrt{{\left(d+D\right)}^2+{\left(\infty \right)}^2}-\sqrt{D^2+{\left(\infty \right)}^2}}{\lambda }\\ & =& 0\end{array}$

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

(b) Calculate the multiple of the λ  that gives the minimum phase difference

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 $1/\lambda$. Therefore, the multiple of the $\lambda$ is zero.

Hence the multiple of the $\lambda$ that gives the minimum phase difference is zero.

(c) Calculate the value yp  at which rays have maximum possible phase difference:

Recall the equation (2).

$\Phi =\frac{\sqrt{{\left(d+D\right)}^2+y_p^2}-\sqrt{D^2+y_p^2}}{\lambda }$

Differentiate the above equation with respect to $y_p$.

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

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

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

From the above equation,

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

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

(d) Calculate the maximum phase difference

Calculate the maximum phase difference.

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

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

Substitute $6\lambda$ for d into above equation.

$\begin{array}{rcl}\Phi & =& \frac{6\lambda }{\lambda }\\ & =& 6\end{array}$

Hence, the maximum phase difference is 6.

(e) Calculate the phase difference when  yp=d

Calculate the phase difference at $y_p=d$.

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

$\Phi =\frac{\sqrt{{\left(d+D\right)}^2+d^2}-\sqrt{D^2+d^2}}{\lambda }$

Substitute $6\lambda$ for d and $20\lambda$ for D into above equation.

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

Solve further as,

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

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

(f) Calculate the intensity at  P when yp=d

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.