Chapter 37, Problem 61AP

(a)

To determine

The longest wavelength that interfere constructively.

Answer to Problem 61AP

The longest wavelength that interfere constructively is $\underset{\_}{72.0\text{m}}$.

Explanation of Solution

Write the expression to find the distance that the ray travels from the top of the transmitter to the ground.

    $x=\sqrt{h^2+{\left(\frac{d}{2}\right)}^2}$                                                                                                        (I)

Here, $h$ is the height above the ground, and $d$ is the distance between the receivers.

Write the expression for the phase shift between the two waves.

    $\delta =2x+\frac{\lambda }{2}-d$                                                                                                         (II)

Here, $\lambda$ is the wavelength, and $x$ is the distance that the ray travels from the top of the transmitter to the ground.

Conclusion:

Substitute $35.0\text{m}$ for $h$ and $50.0\text{m}$ for $d$ in equation (I).

    $\begin{array}{c}x=\sqrt{{\left(35.0\text{m}\right)}^2+{\left(\frac{50.0\text{m}}{2}\right)}^2}\\ =\sqrt{1850{\text{m}}^2}\\ =43.0\text{m}\end{array}$

Substitute $m\lambda$ for $\delta$ in equation (II).

    $m\lambda =2x+\frac{\lambda }{2}-d$

Rewrite the above equation to find the wavelength.

    $\lambda =\frac{2x-d}{\left(m-\frac{1}{2}\right)}$

Substitute $1$ for $m$, $43.0\text{m}$ for $x$ and $50.0\text{m}$ for $d$ in the above equation.

    $\begin{array}{c}\lambda =\frac{2\left(43.0\text{m}\right)-50.0\text{m}}{\left(1-\frac{1}{2}\right)}\\ =72.0\text{m}\end{array}$

Therefore, the longest wavelength that interfere constructively is $\underset{\_}{72.0\text{m}}$.

(b)

To determine

The longest wavelength that interfere destructively .

Answer to Problem 61AP

The longest wavelength that interfere destructively is $\underset{\_}{36.0\text{m}}$.

Explanation of Solution

Write the expression to find the distance that the ray travels from the top of the transmitter to the ground.

    $x=\sqrt{h^2+{\left(\frac{d}{2}\right)}^2}$                                                                                                        (I)

Here, $h$ is the height above the ground and $d$ is the distance between the receivers.

Write the expression for the phase shift between the two waves.

    $\delta =2x+\frac{\lambda }{2}-d$                                                                                                         (II)

Here, $\lambda$ is the wavelength and $x$ is the distance that the ray travels from the top of the transmitter to the ground.

Conclusion:

Substitute $35.0\text{m}$ for $h$ and $50.0\text{m}$ for $d$ in equation (I).

    $\begin{array}{c}x=\sqrt{{\left(35.0\text{m}\right)}^2+{\left(\frac{50.0\text{m}}{2}\right)}^2}\\ =\sqrt{1850{\text{m}}^2}\\ =43.0\text{m}\end{array}$

Substitute $\left(m+\frac{1}{2}\right)\lambda$ for $\delta$ in equation (II).

    $\left(m+\frac{1}{2}\right)\lambda =2x+\frac{\lambda }{2}-d$

Rewrite the above equation to find the wavelength.

    $\lambda =\frac{2x-d}{m}$

Substitute $1$ for $m$, $43.0\text{m}$ for $x$ and $50.0\text{m}$ for $d$ in the above equation.

    $\begin{array}{c}\lambda =\frac{2\left(43.0\text{m}\right)-50.0\text{m}}{1}\\ =36.0\text{m}\end{array}$

Therefore, the longest wavelength that interfere destructively is $\underset{\_}{36.0\text{m}}$.