Question

In Fig. 17-25, two point sources $\mathbf{S}\mathbf{1}$and$\mathbf{S}\mathbf{2}$, which are in phase, emitidentical sound waves of wavelength$\mathbf{2}\mathbf{.}\mathbf{0}\mathbf{}\mathit{m}$. In terms of wavelengths, what is the phase differencebetween the waves arriving atpoint Pif (a)$\mathbf{L}\mathbf{1}\mathbf{}\mathbf{=}\mathbf{38}\mathbf{}\mathit{m}\mathbf{}$and$\mathbf{L}\mathbf{2}\mathbf{}\mathbf{=}\mathbf{}\mathbf{34}\mathbf{}\mathit{m}$, and (b)$\mathbf{L}\mathbf{1}\mathbf{}\mathbf{=}\mathbf{39}\mathbf{}\mathit{m}$and$\mathbf{L}\mathbf{2}\mathbf{}\mathbf{=}\mathbf{}\mathbf{36}\mathbf{}\mathit{m}$? (c) Assuming that the source separation is much smaller than$\mathbf{L}\mathbf{1}$and$\mathbf{L}\mathbf{2}$, what type of interference occurs atin situations (a) and (b)?

Short answer

1. The phase difference between the points arriving at P if$L_1=38m$ and$L_2=34m$ is$4\mathrm{\pi }$
2. The phase difference between the points arriving at P if $L_1=39m$and $L_2=36m$is$3\mathrm{\pi }$
3. Type of interference at P in situation (a) is constructive interference
4. Type of interference at P in situation (b) is destructive interference

Step-by-step solution

Step 1: Given

1. The wavelength of the two in phase waves =$\lambda =2.0m$
2. In situation (a), the distance of point P from the source are$L_1=38m,L_2=34m$
3. In situation (b), the distance of point P from the source are$L_1=39m,L_2=36m$

Determining the concept

The interference between two waves with identical wavelengths occurs when they cross at a point. Constructive interference is observed at that point if the phase difference is an integral multiple of 2π and destructive interference is observed if the phase difference is an odd integral multiple of π. The phase difference is decided by the path difference between the two waves.

Formulae are as follows:

$f=\frac{∆L}{\lambda }2\mathrm{\pi }$

Condition for constructive interference$f=m\left(2\pi \right)\dots \dots \dots \dots form=0,1,2,\dots .$

Condition for destructive interference $f=\left(2m+1\right)\pi \dots \dots .form=0,1,2,\dots .$

Where, L is distance from source,$\lambda$is wavelength.

(a) Determining thephase difference between the points arriving at P if  L1=38 mand L2=34 m

Determine the phase difference between the waves as,

$\Delta L=L_1-L_2=38-34=4.0m$

Using the formula,

$f=\frac{∆L}{\lambda }2\mathrm{\pi }$

$\therefore f=\frac{4.0}{2.0}\times 2\pi$

$\therefore f=4\pi$

Hence,the phase difference between the points arriving at P if$L_1=38m$ and $L_2=34m$is$4\mathrm{\pi }$

 Step 4: (b) Determining the phase difference between the points arriving at P if L1=39mand L2=36m

Determine the phase difference between the waves as ,

$\Delta L=L_1-L_2=39-36=3.0m$

Using the formula,

$f=\frac{∆L}{\lambda }2\mathrm{\pi }$

$\therefore f=\frac{3.0}{2.0}\times 2\pi$

$\therefore f=3\pi$

Hence, the phase difference between the points arriving at P if $L_1=39m$and $L_2=36m$is$3\mathrm{\pi }$

(c) Determining the type of interference at P in situation (a) 

In part (a), it is seen that the phase difference is an integer multiple of $2\mathrm{\pi }$hence, the interference will be constructive.

Hence, type of interference at P in situation (a) is constructive interference

(d) determining the type of interference at P in situation (b) 

In part (b), the phase difference is an odd integer multiple of. Hence, the interference will be destructive.

Hence, type of interference at P in situation (b) is destructive interference

Therefore, the two waves having identical wavelengths interfere when they cross a point at the same instant. The type of interference is decided by the phase difference between the waves at that point. Hence, we calculate the phase difference to determine the type.