Chapter 24, Problem 42P

(a)

To determine

The number of directions on the other side of the array in which there is a maximum of intensity.

Answer to Problem 42P

The number of directions on the other side of the array in which there is a maximum of intensity is three corresponding to the orders -1, 0 and +1.

Explanation of Solution

Given Info: The velocity of light is $343\text{m/s}$, the frequency is $37.2\text{kHz}$, the slit separation is $1.30\text{cm}$ and the maximum viewing angle is $90.0°$.

Formula to calculate the order is,

$\left|m\right|\le \left(\frac{d}{\lambda }\right)\sin \left|\theta \right|$

• $\lambda$ is the wavelength
• $m$ is the order
• $d$ is the grid separation
• $\theta$ is the angle
• $v$ is the velocity of sound
• $f$ is the frequency

Use $v/f$ for $\lambda$ and rearrange the above equation.

$\left|m\right|\le \left(\frac{d}{v/f}\right)\sin \left|\theta \right|$

Substitute $343\text{m/s}$ for $v$, $37.2\text{kHz}$ for $f$, $1.30\text{cm}$ for $d$ and $90.0°$ for $\theta$ to find $\left|m\right|$.

$\begin{array}{c}\left|m\right|\le \left(\frac{\left(\left(1.30\text{cm}\right)\left(\frac{1\text{m}}{{10}^2\text{cm}}\right)\right)}{343\text{m/s}/\left(\left(37.2\text{kHz}\right)\left(\frac{{10}^3\text{Hz}}{1\text{kHz}}\right)\right)}\right)\sin \left|90.0°\right|\\ \le 1.41\end{array}$

Thus, the number of directions on the other side of the array in which there is a maximum of intensity is three corresponding to the orders -1, 0 and +1.

Conclusion:

The number of directions on the other side of the array in which there is a maximum of intensity is three corresponding to the orders -1, 0 and +1.

(b)

To determine

The angle for each of the directions relative to the direction of incident beam.

Answer to Problem 42P

The angles for each of the directions relative to the direction of incident beam are $-45.2°,0°\text{and}45.2°$ for orders -1, 0 and -1 respectively.

Explanation of Solution

Given Info: The velocity of light is $343\text{m/s}$, the frequency is $37.2\text{kHz}$, the slit separation is $1.30\text{cm}$ and the maximum viewing angle is $90.0°$.

Formula to calculate the angle is,

$\begin{array}{c}\theta ={\sin }^{-1}\left(\frac{m\lambda }{d}\right)\\ ={\sin }^{-1}\left(\frac{m\left(v/f\right)}{d}\right)\end{array}$

Substitute $343\text{m/s}$ for $v$, $37.2\text{kHz}$ for $f$, $1.30\text{cm}$ for $d$ and $-1$ for $m$ to find $\theta$.

$\begin{array}{c}\theta ={\sin }^{-1}\left(\frac{\left(-1\right)\left(343\text{m/s}/\left(\left(37.2\text{kHz}\right)\left(\frac{{10}^3\text{Hz}}{1\text{kHz}}\right)\right)\right)}{\left(\left(1.30\text{cm}\right)\left(\frac{1\text{m}}{{10}^2\text{cm}}\right)\right)}\right)\\ ={\sin }^{-1}\left(-0.709\right)\\ =-45.2°\end{array}$

Substitute $343\text{m/s}$ for $v$, $37.2\text{kHz}$ for $f$, $1.30\text{cm}$ for $d$ and $0$ for $m$ to find $\theta$.

$\begin{array}{c}\theta ={\sin }^{-1}\left(\frac{\left(0\right)\left(343\text{m/s}/\left(\left(37.2\text{kHz}\right)\left(\frac{{10}^3\text{Hz}}{1\text{kHz}}\right)\right)\right)}{\left(\left(1.30\text{cm}\right)\left(\frac{1\text{m}}{{10}^2\text{cm}}\right)\right)}\right)\\ =0°\end{array}$

Substitute $343\text{m/s}$ for $v$, $37.2\text{kHz}$ for $f$, $1.30\text{cm}$ for $d$ and $1$ for $m$ to find $\theta$.

$\begin{array}{c}\theta ={\sin }^{-1}\left(\frac{\left(1\right)\left(343\text{m/s}/\left(\left(37.2\text{kHz}\right)\left(\frac{{10}^3\text{Hz}}{1\text{kHz}}\right)\right)\right)}{\left(\left(1.30\text{cm}\right)\left(\frac{1\text{m}}{{10}^2\text{cm}}\right)\right)}\right)\\ ={\sin }^{-1}\left(0.709\right)\\ =+45.2°\end{array}$

Thus, the angles for each of the directions relative to the direction of incident beam are $-45.2°,0°\text{and}+45.2°$ for orders -1, 0 and -1 respectively.

Conclusion:

The angles for each of the directions relative to the direction of incident beam are $-45.2°,0°\text{and}+45.2°$ for orders -1, 0 and -1 respectively.