Chapter 25, Problem 61P

(a)

To determine

Whether the power detected at $\theta =0$ a maximum or minimum.

Answer to Problem 61P

The power detected at $\theta =0°$ is a maximum.

Explanation of Solution

It is given that the radio waves have the same frequency and start out in phase. This implies the radio waves are coherent. When two coherent waves are in phase, their superposition will result in constructive interference.

All the points on the $\theta =0°$ line are equidistant from the towers. Since the waves started out in phase, they will arrive in phase and will combine constructively at all points on $\theta =0°$ . Since the radio waves interfere constructively at $\theta =0°$ , the power detected at $\theta =0°$ will be maximum.

(b)

To determine

The graph of $P$ versus $\theta$ to show the variation of power with the angle and to label the graphs with values of $\theta$ at which power is maximum or minimum.

Answer to Problem 61P

The qualitative graph showing the variation of power with the angle is

Explanation of Solution

The given situation can be treated as a double-slit problem with the antennas playing the role of the slits.

Write the condition for the maxima.

  $d\sin \theta =m\lambda$                                                                                                             (I)

Here, $d$ is the distance between the antennas, $\theta$ is the angle, $m$ is the order of the fringe and $\lambda$ is the wavelength of the light.

It is given that $d=\lambda$ .

Replace $d$ in equation (I) by $\lambda$ and rewrite the equation for $\theta$ .

  $\begin{array}{c}\lambda \sin \theta =m\lambda \\ \theta ={\sin }^{-1}m\end{array}$                                                                                                   (II)

The values of $0$ and $\pm 1$ for $m$ give maxima for the directions between $-90°$ and $+90°$ .

Substitute $0$ for $m$ in equation (II) to find the corresponding angle.

  $\begin{array}{c}\theta ={\sin }^{-1}\left(0\right)\\ =0\end{array}$

Substitute $\pm 1$ for $m$ in equation (II) to find the corresponding angle.

  $\begin{array}{c}\theta ={\sin }^{-1}\left(\pm 1\right)\\ =\pm 90°\end{array}$

Thus the directions between $-90°$ and $+90°$ for interference maxima are $0°$ and $\pm 90°$ . The directions of maximum interference for $90°<\theta <180°$ and $-180°<\theta <-90°$ will be the mirror image of values found above. This implies for $-180°<\theta <180°$ , maxima occur at $\theta =-180°,-90°,0°,90°\text{ and }180°$ .

Write the equation for the minima.

  $d\sin \theta =\left(m+\frac{1}{2}\right)\lambda$                                                                                              (III)

Replace $d$ in equation (III) by $\lambda$ and rewrite the equation for $\theta$ .

  $\begin{array}{c}\lambda \sin \theta =\left(m+\frac{1}{2}\right)\lambda \\ \theta ={\sin }^{-1}\left(m+\frac{1}{2}\right)\end{array}$                                                                                     (IV)

Substitute $0$ for $m$ in equation (IV) to find the corresponding angle.

    $\begin{array}{c}\theta ={\sin }^{-1}\left(0+\frac{1}{2}\right)\\ ={\sin }^{-1}\frac{1}{2}\\ =30°\end{array}$

$1$ cannot be substituted for $m$ in equation (IV) since there is no angle whose sine is $1.5$ . This implies the only directions between $-90°$ and $+90°$ where interference minimum occur are $-30°$ and $+30°$ . The directions of minimum interference for $90°<\theta <180°$ and $-180°<\theta <-90°$ will be the mirror image of values found above. This implies for $-180°<\theta <180°$ , minima occur at $\theta =\pm 30°\text{ and }\pm 150°$ .

The qualitative graph is shown in figure 1.

Conclusion:

Thus, the graph of $P$ versus $\theta$ to showing the variation of power with the angle is shown in figure 1 and values of $\theta$ at which power is maximum or minimum are labelled.

(c)

To determine

The graph of $P$ versus $\theta$ for $d=\lambda /2$ to show the variation of power with the angle and to label the graphs with values of $\theta$ at which power is maximum or minimum.

Answer to Problem 61P

The qualitative graph for $d=\lambda /2$ showing the variation of power with the angle is

Explanation of Solution

Replace $d$ in equation (I) by $\lambda /2$ and rewrite the equation for $\theta$ .

  $\begin{array}{c}\frac{\lambda }{2}\sin \theta =m\lambda \\ \sin \theta =2m\\ \theta ={\sin }^{-1}\left(2m\right)\end{array}$ (V)

Substitute $0$ for $m$ in equation (V) to find the corresponding angle.

  $\begin{array}{c}\theta ={\sin }^{-1}\left(0\right)\\ =0\end{array}$

$1$ cannot be substituted for $m$ in equation (V) since there is no angle whose sine is $2.0$ . Considering the symmetry as in part (b), $\theta =\pm 180°$ will also yield maxima.

Replace $d$ in equation (III) by $\lambda /2$ and rewrite the equation for $\theta$ .

  $\begin{array}{c}\frac{\lambda }{2}\sin \theta =\left(m+\frac{1}{2}\right)\lambda \\ \sin \theta =2m+1\\ \theta ={\sin }^{-1}\left(2m+1\right)\end{array}$ (VI)

Substitute $0$ for $m$ in equation (IV) to find the corresponding angle where minima occur.

  $\begin{array}{c}\theta ={\sin }^{-1}\left(0+1\right)\\ =90°\end{array}$

$1$ cannot be substituted for $m$ in equation (VI) since there is no angle whose sine is $3.0$ . This implies the only directions between $-180°<\theta <180°$ where minima occur are $\theta =\pm 90°$ .

The qualitative graph is shown in figure 2.

Conclusion:

Thus, the graph of $P$ versus $\theta$ to for $d=\lambda /2$ showing the variation of power with the angle is shown in figure 2 and values of $\theta$ at which power is maximum or minimum are labelled.