Chapter 25, Problem 93P

(a)

To determine

The resultant intensity when the waves interfere constructively.

Answer to Problem 93P

The resultant intensity when the waves interfere constructively is $\underset{\_}{0.191{\text{W/m}}^2}$.

Explanation of Solution

Given that the amplitudes of the interfering waves are $3.0\text{V/m}$ and $9.0\text{V/m}$.

Write the expression for the intensity of the EM wave.

  $I=\langle u\rangle c$                                                                                                                (I)

Here, $I$ is the intensity, $\langle u\rangle$ is the average energy density of the wave, and $c$ is the speed of light.

Write the expression for the average energy density of the wave.

  $\langle u\rangle ={\epsilon }_0{E}_{\text{rms}}^2$                                                                                                          (II)

Here, ${\epsilon }_0$ is the permittivity of free space, and $E_{\text{rms}}$ is the rms value of the amplitude of the wave.

Write the expression for the rms amplitude of the wave.

  $E_{\text{rms}}=\frac{E_{\text{m}}}{\sqrt{2}}$                                                                                                         (III)

Here, $E_{\text{m}}$ is the maximum amplitude.

Use equation (III) in (II).

  $\begin{array}{c}\langle u\rangle ={\epsilon }_0{\left(\frac{E_{\text{m}}}{\sqrt{2}}\right)}^2\\ =\frac{1}{2}{\epsilon }_0{E}_{\text{m}}^2\end{array}$                                                                                                     (IV)

Use equation (IV) in (I).

  $I=\frac{1}{2}{\epsilon }_{0}c{E}_{\text{m}}^2$                                                                                                    (V)

Write the expression for the resultant amplitude when two EM waves undergo constructive interference.

  $E_{\text{m}}=E_{\text{m1}}+E_{\text{m2}}$                                                                                               (VI)

Conclusion:

Substitute $3.0\text{V/m}$ for $E_{\text{m1}}$ and $9.0\text{V/m}$ for $E_{\text{m2}}$ in equation (VI) to find $E_{\text{m}}$.

  $\begin{array}{c}{E}_{\text{m}}=3.0\text{V/m}+9.0\text{V/m}\\ =12.0\text{V/m}\end{array}$

Substitute $12.0\text{V/m}$ for $E_{\text{m}}$, $3.00\times {10}^8\text{m/s}$ for, $8.85\times {10}^{-12}{\text{C}}^2\text{/N}\cdot {\text{m}}^{\text{2}}$ for ${\epsilon }_0$ in equation (V) to find $I$.

  $\begin{array}{c}I=\frac{1}{2}\left(8.85\times {10}^{-12}{\text{C}}^2\text{/N}\cdot {\text{m}}^{\text{2}}\right)\left(3.00\times {10}^8\text{m/s}\right){\left(12.0\text{V/m}\right)}^2\\ =0.191{\text{W/m}}^2\end{array}$

Therefore, the resultant intensity when the waves interfere constructively is $\underset{\_}{0.191{\text{W/m}}^2}$.

(b)

To determine

The resultant intensity when the waves interfere destructively.

Answer to Problem 93P

The resultant intensity when the waves interfere destructively is $\underset{\_}{0.048{\text{W/m}}^2}$.

Explanation of Solution

Given that the amplitudes of the interfering waves are $3.0\text{V/m}$ and $9.0\text{V/m}$.

Write the expression for the resultant amplitude when two EM waves undergo destructive interference.

  $E_{\text{m}}=\left|E_{\text{m1}}-E_{\text{m2}}\right|$                                                                                                 (VII)

Equation (V) fives the resultant intensity of the wave.

  $I=\frac{1}{2}{\epsilon }_{0}c{E}_{\text{m}}^2$

Conclusion:

Substitute $3.0\text{V/m}$ for $E_{\text{m1}}$ and $9.0\text{V/m}$ for $E_{\text{m2}}$ in equation (VII) to find $E_{\text{m}}$.

  $\begin{array}{c}{E}_{\text{m}}=\left|3.0\text{V/m}-9.0\text{V/m}\right|\\ =6.0\text{V/m}\end{array}$

Substitute $6.0\text{V/m}$ for $E_{\text{m}}$, $3.00\times {10}^8\text{m/s}$ for, $8.85\times {10}^{-12}{\text{C}}^2\text{/N}\cdot {\text{m}}^{\text{2}}$ for ${\epsilon }_0$ in equation (V) to find $I$.

  $\begin{array}{c}I=\frac{1}{2}\left(8.85\times {10}^{-12}{\text{C}}^2\text{/N}\cdot {\text{m}}^{\text{2}}\right)\left(3.00\times {10}^8\text{m/s}\right){\left(6.0\text{V/m}\right)}^2\\ =0.048{\text{W/m}}^2\end{array}$

Therefore, the resultant intensity when the waves interfere destructively is $\underset{\_}{0.048{\text{W/m}}^2}$.