Chapter 38, Problem 27P

Consider an array of parallel wires with uniform spacing of 1.30 cm between centers. In air at 20.0°C, ultrasound with a frequency of 37.2 kHz from a distant source is incident perpendicular to the array. (a) Find the number of directions on the other side of the array in which there is a maximum of intensity. (b) Find the angle for each of these directions relative to the direction of the incident beam.

To determine

The number of directions on the other side of the array for maximum intensity.

Answer to Problem 27P

The number of directions on the other side of the array for maximum intensity is three.

Explanation of Solution

Given info: Temperature of air is $20.0°\text{C}$, spacing between centre is $1.30\text{cm}$ and frequency of array is $37.2\text{kHz}$.

The wavelength for a diffraction grating can be given as,

$\lambda =\frac{v}{f}$

Here,

$\lambda$ is the wavelength of light.

$v$ is the speed of sound.

$f$ is the frequency of the array.

Substitute $343\text{m}/\text{s}$ for $v$ and $37.2\text{kHz}$ for $f$ in the above equation to find $\lambda$,

$\begin{array}{c}\lambda =\frac{\left(343\text{m}/\text{s}\right)}{\left(37.2\text{kHz}\right)\left(\frac{{10}^3\text{Hz}}{1\text{kHz}}\right)}\\ =9.22\times {10}^{-3}\text{m}\end{array}$

The condition for the bright fringe in diffraction can be given as,

$m\lambda =d\sin \theta$ (1)

Here,

$\theta$ is the angle of spectral line.

$\lambda$ is the wavelength of light.

$m$ is the order of diffraction.

$d$ is the spacing between centre.

Substitute $90°$ for $\theta$, $9.22\times {10}^{-3}\text{m}$ for $\lambda$ and $1.30\text{cm}$ for $d$ in the equation (1),

$\begin{array}{c}m\left(9.22\times {10}^{-3}\text{m}\right)=\left[\left(1.30\text{cm}\right)\left(\frac{1\text{m}}{100\text{cm}}\right)\right]\sin 90°\\ m=1.41\\ \approx 1\end{array}$

The maximum number of direction possible can be given as,

$m_{\max }=2m+1$

Here,

$m_{\max }$ is the maximum number of directions.

Substitute $1$ for $m$ in the above equation,

$\begin{array}{c}{m}_{\max }=2(1)+1\\ =3\end{array}$

Thus, the number of directions on the other side of the array for maximum intensity is three.

Conclusion:

Therefore, the number of directions on the other side of the array for maximum intensity is three.

To determine

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

Answer to Problem 27P

The angle for each of the directions relative to the direction of the incident beam is $0°$, $+45.2°$ and $-45.2°$.

Explanation of Solution

Given info: Temperature of air is $20.0°\text{C}$, spacing between centre is $1.30\text{cm}$ and frequency of array is $37.2\text{kHz}$.

The condition for a diffraction grating as in equation (1) can be given as,

$m\lambda =d\sin \theta$

Rearrange the above expression for $\theta$,

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

Substitute $\left(-2\right)$ for $m$, $9.22\times {10}^{-3}\text{m}$ for $\lambda$ and $1.30\text{cm}$ for $d$ in the equation (2),

$\begin{array}{c}\theta ={\sin }^{-1}\left(\frac{\left(-2\right)\left(9.22\times {10}^{-3}\text{m}\right)}{\left(1.30\text{cm}\right)\left(\frac{1\text{m}}{100\text{cm}}\right)}\right)\\ ={\sin }^{-1}\left(-1.418\right)\end{array}$

As the range of sine function is $\left[-1,1\right]$, the value of ${\sin }^{-1}\left(-1.418\right)$ is undetermined. Therefore, $\left|m\right|$ cannot be more than unity.

Substitute $\left(-1\right)$ for $m$, $9.22\times {10}^{-3}\text{m}$ for $\lambda$ and $1.30\text{cm}$ for $d$ in the equation (2),

$\begin{array}{c}\theta ={\sin }^{-1}\left(\frac{\left(-1\right)\left(9.22\times {10}^{-3}\text{m}\right)}{\left(1.30\text{cm}\right)\left(\frac{1\text{m}}{100\text{cm}}\right)}\right)\\ ={\sin }^{-1}\left(-0.709\right)\\ =-45.2°\end{array}$

Thus, $\theta$ is $\left(-45.2°\right)$ for the $\left(-1\right)$ order of diffraction.

Substitute $0$ for $m$, $9.22\times {10}^{-3}\text{m}$ for $\lambda$ and $1.30\text{cm}$ for $d$ in the equation (2),

$\begin{array}{c}\theta ={\sin }^{-1}\left(\frac{\left(0\right)\left(9.22\times {10}^{-3}\text{m}\right)}{\left(1.30\text{cm}\right)\left(\frac{1\text{m}}{100\text{cm}}\right)}\right)\\ ={\sin }^{-1}\left(0\right)\\ =0°\end{array}$

Thus, $\theta$ is $0°$ for the $0$ order of diffraction.

Substitute $\left(1\right)$ for $m$, $9.22\times {10}^{-3}\text{m}$ for $\lambda$ and $1.30\text{cm}$ for $d$ in the equation (2),

$\begin{array}{c}\theta ={\sin }^{-1}\left(\frac{\left(1\right)\left(9.22\times {10}^{-3}\text{m}\right)}{\left(1.30\text{cm}\right)\left(\frac{1\text{m}}{100\text{cm}}\right)}\right)\\ ={\sin }^{-1}\left(0.709\right)\\ =45.2°\end{array}$

Thus, $\theta$ is $45.2°$ for the $\left(1\right)$ order of diffraction.

Conclusion:

Therefore, the angle for each of the directions relative to the direction of the incident beam is $0°$, $+45.2°$ and $-45.2°$.