Question

FIGURE EX17.27 shows the circular wave fronts emitted by two

wave sources.

a. Are these sources in phase or out of phase? Explain.

b. Make a table with rows labeled P, Q, and R and columns

labeled $r_1,r_2,∆r,$and C/D. Fill in the table for points P, Q, and

R, giving the distances as multiples of l and indicating, with a

C or a D, whether the interference at that point is constructive

or destructive.

Short answer

The two wavefronts produce destructive interference at point Q and constructive interference at point P and R.

Step-by-step solution

Part a Step 1: Introduction

When two wavefronts are said to be moving out of phase, they are not equidistant from the source.

When two wavefronts travel equidistance from the source which means the wavefronts travels same distance in a period of time, then the wavefronts are said to be in phase.

Explanation

The circular wavefronts emitted by the two sources show that the two sources are in phase because the wavefronts of each source have moved the same distance from their sources.

Part b Step 1: Determination of the type of interference

• According to the given diagram, the wavefront labelled as 1 and 2 are moving same distance. So, both of the sources are in phase. Thus, the phase change is ${\varphi }_o=0$
• At point P, the wavefront 1 moves a distance of $r_1=3\lambda$and wavefront 2 moves a distance of $r_2=4\lambda$.Therefore, the phase difference of the two waves at the point is,

$$\begin{gathered} ∆\varphi =\frac{2\pi ∆r}{\lambda }+{\varphi }_o \\ ∆\varphi =\frac{2\pi \left(4\lambda -3\lambda \right)}{\lambda }+{\varphi }_o \\ ∆\varphi =2\pi +0 \\ ∆\varphi =2\pi \end{gathered}$$

This corresponds to constructive interference.

• At point Q, the wavefront 1 moves a distance of $r_1=\frac{7\lambda }{2}$and wavefront 2 moves a distance of $r_2=2\lambda$. Therefore, the phase difference of the two waves at the point is,

role="math" localid="1650373377486" $$\begin{gathered} ∆\varphi =\frac{2\pi ∆r}{\lambda }+{\varphi }_o \\ ∆\varphi =\frac{2\pi \left({\displaystyle \frac{7\lambda }{2}}-{\displaystyle 2\lambda }\right)}{\lambda }+0 \\ ∆\varphi =\frac{\left(2\pi \times \frac{3\lambda }{2}\right)}{\lambda }+0 \\ ∆\varphi =3\pi +0 \\ ∆\varphi =3\pi \end{gathered}$$

This corresponds to destructive interference.

• At point R, the wavefront 1 moves a distance of $r_1=\frac{5\lambda }{2}$and wavefront 2 moves a distance of $r_2=\frac{7\lambda }{2}$. Therefore, the phase difference of the two waves at the point is,

$$\begin{gathered} ∆\varphi =\frac{2\pi ∆r}{\lambda }+{\varphi }_o \\ ∆\varphi =\frac{2\pi \left({\displaystyle \frac{7\lambda }{2}-\frac{5\lambda }{2}}\right)}{\lambda }+0 \\ ∆\varphi =\frac{\left(2\pi \times \lambda \right)}{\lambda }+0 \\ ∆\varphi =2\pi +0 \\ ∆\varphi =2\pi \end{gathered}$$

This corresponds to constructive interference.

Table of interference pattern

If we mention the details of the interference in a table, it should be like:

				C/D
P	$3\lambda$	$4\lambda$	$\lambda$	C
Q	$\frac{7\lambda }{2}$	$2\lambda$	$\frac{3\lambda }{2}$	D
R		$\frac{7\lambda }{2}$	$\lambda$	C