Chapter 27, Problem 45P

(a)

To determine

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

Answer to Problem 45P

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.

(b)

To determine

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

Answer to Problem 45P

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 $-45.2°$.