Question

In Fig. 17-45, sound waves$\mathbf{A}$and$\mathbf{B}$, both of wavelength$\mathbf{\lambda }$, are initially in phase and traveling rightward, as indicated by the two rays. Wave Ais reflected from four surfaces but ends up traveling in its original direction. What multiple of wavelength$\mathbf{\lambda }$is the smallest value of distance$\mathbf{L}$in the figure that puts Aand Bexactly out of phase with each other after the reflections?

Short answer

The multiple of wavelength, which is the smallest value of distance, is$0.25$

Step-by-step solution

The given data

Two waves are out of phase.

Understanding the concept of interference

Here we can use the concept of interference. When two waves are out of phase, the phase difference between them is $\mathbf{\pi }$. We can find the multiple wavelengths using this concept.

Formula:

The phase of a wave,

$\mathbf{\phi }\mathbf{=}\frac{\mathbf{\Delta L}}{\mathbf{\lambda }}\mathbf{2}\mathbf{\pi }$ …(i)

Calculation of the multiple of wavelength

Here, the additional length travelled by wave A is vertical zig zag, and that is equal to $2\mathrm{L}$.

Here we are given that waves are out of phase. Hence, the phase angle using equation (i) and given values is given as:$\mathrm{\pi }=\frac{2\mathrm{L}}{\mathrm{\lambda }}2\mathrm{\pi }$

Also the path difference should be equal to$\mathrm{\pi }$since the waves are out of phase.

$\begin{array}{c}2\mathrm{L}=\frac{\mathrm{\lambda }}{2}\\ \frac{\mathrm{L}}{\mathrm{\lambda }}=\frac{1}{4}\\ \frac{\mathrm{L}}{\mathrm{\lambda }}=0.25\\ \mathrm{L}=0.25\mathrm{\lambda }\end{array}$

Hence, the value of the multiple is $0.25$.