Chapter 35, Problem 1OQ

To determine

The rank for the given situations.

Answer to Problem 1OQ

The ranking order for the given situation is $\left(\text{e}\right)>\left(\text{c}\right)>\left(\text{b}\right)>\left(\text{a}\right)>\left(\text{d}\right)$.

Explanation of Solution

The smaller the ratio of wavelength and the distance travelled by the waves, the better is the wave approximation.

    $\frac{\lambda }{d}$

Here, $\lambda$ is the wavelength and $d$ is the distance travelled.

Conclusion:

Option (a): The sound of a low whistle at $1\text{kHz}$ passes through a doorway $1\text{m}$ wide.

The wavelength for the low whistle is approximately $0.34\text{m}$.

Substitute $0.34\text{m}$ for $\lambda$ and $1\text{m}$ for $d$ in equation (I) to solve for $\frac{\lambda }{d}$.

    $\begin{array}{c}\frac{\lambda }{d}=\frac{0.34\text{m}}{1\text{m}}\\ =0.34\end{array}$

Option (b): Red light passes through the pupil of your eye.

The wavelength of red light is $0.7\times {10}^{-6}\text{m}$ and the width of the pupil of the human eye is approximately $2\text{mm}$.

Substitute $0.7\times {10}^{-6}\text{m}$ for $\lambda$ and $2\text{mm}$ for $d$ in equation (I) to solve for $\frac{\lambda }{d}$.

    $\begin{array}{c}\frac{\lambda }{d}=\frac{0.7\times {10}^{-6}\text{m}}{2\text{mm}\times \frac{{10}^{-3}\text{m}}{1\text{mm}}}\\ =3.5\times {10}^{-4}\end{array}$

Option (c): Blue light passes through the pupil of your eye.

The wavelength of blue light is $0.4\times 10^{-6}\text{m}$.

Substitute $0.4\times 10^{-6}\text{m}$ for $\lambda$ and $2\text{mm}$ for $d$ in equation (I) to solve for $\frac{\lambda }{d}$.

    $\begin{array}{c}\frac{\lambda }{d}=\frac{0.4\times 10^{-6}\text{m}}{2\text{mm}\times \frac{{10}^{-3}\text{m}}{1\text{mm}}}\\ =2\times {10}^{-4}\end{array}$

Option (d): The wave broadcast by an AM radio station passes through a doorway 1 m wide.

The wavelength of the of the wave broadcast by the AM radio station is $300\text{m}$.

Substitute $300\text{m}$ for $\lambda$ and $1\text{m}$ for $d$ in equation (I) to solve for $\frac{\lambda }{d}$.

    $\begin{array}{c}\frac{\lambda }{d}=\frac{300\text{m}}{1\text{m}}\\ =300\end{array}$

Option (e): An x-ray passes through the space between bones in your elbow joint.

The wavelength of x-ray is $1\times {10}^{-9}\text{m}$ and distance is $1\text{mm}$.

    $\begin{array}{c}\frac{\lambda }{d}=\frac{1\times {10}^{-9}\text{m}}{1\text{mm}\times \frac{{10}^{-3}\text{m}}{1\text{mm}}}\\ =0.000001\end{array}$

Therefore, the ranking order for the given situation is $\left(\text{e}\right)>\left(\text{c}\right)>\left(\text{b}\right)>\left(\text{a}\right)>\left(\text{d}\right)$.