Question

$S_1$ and $S_2$ are two sources of sound emitting sine waves. The two sources are in phase. The sound emitted by the two sources interfere at point $F$. The waves of wavelength:

• A. $\frac{1}{2}\text{m}$ will result in constructive interference
• B. $\frac{3}{2}\text{m}$ will result in constructive interference
• C. $2\text{m}$ will result in destructive interference
• D. $4\text{m}$ will result in destructive interference

Answer 1

The correct option is D $4\text{m}$ will result in destructive interference

we know that,
for constructive interference the path difference $=n\lambda$
$S_{1}F-S_{2}F=n\lambda$
$6\text{m}-4\text{m}=n\lambda$
$2\text{m}=n\lambda$
$\lambda =\frac{2}{n}\text{m}$
where $n=1,2,...$
So, $\lambda =2\text{m},1\text{m},\frac{2}{3}\text{m},...$
Hence, the wave length is $1\text{m}$ in constuctive interference.

For destructive interference the path difference $=(2n+1)\frac{\lambda }{2}$
$S_{1}F-S_{2}F=(2n+1)\frac{\lambda }{2}$
$6\text{m}-4\text{m}=(2n+1)\frac{\lambda }{2}$
$2\text{m}=(2n+1)\frac{\lambda }{2}$
$\lambda =\frac{4}{2n+1}\text{m}$
where $n=0,1,2,...$
So, $\lambda =4\text{m},\frac{4}{3}\text{m},\frac{4}{5}\text{m},...$
Hence, the wave length is $4\text{m}$ in destuctive interference.