91Ó°ÊÓ

The path difference is found out by individual calculation of the distance travelled by both waves and then subsequent subtraction.

Let P1is the distance between the observer ad the right speaker and P2is the distance between the observer ad the left speaker.

From the figure provided,

${\mathrm{P}}_1=8.0 \mathrm{m}$

$$\begin{gathered} {\mathrm{p}}_2=\sqrt{{\left(6.0{\mathrm{m}}^2+(8.0\mathrm{m}\right)}^2} \\ =10.0\mathrm{m} \end{gathered}$$

Thus, the path difference,

localid="1663927893781" $$\begin{gathered} ∆\mathrm{P}={\mathrm{p}}_2-{\mathrm{P}}_1 \\ =10.0\mathrm{m}-8.0\mathrm{m} \\ =2.0\mathrm{m} \end{gathered}$$

The path difference given as 2.0 m and the wavelength is gives as 2.0 m.

For constructive interference, the condition is,

${\mathrm{r}}_2-{\mathrm{r}}_1=\mathrm{³¾Î»},      \mathrm{m}=0,Â\pm 1,Â\pm 2,...$

Clearly from the given data, m = 1 which makes the path difference an integral multiple of the wavelength. So the interference is constructive.

The distance is increased to 17.0 m. So,

From the figure provided,

${\mathrm{P}}_1=17.0\mathrm{m}$

localid="1663928096577" $$\begin{gathered} {\mathrm{p}}_2=\sqrt{{\left(6.0{\mathrm{m}}^2+(8.0\mathrm{m}\right)}^2} \\ =18.0\mathrm{m} \end{gathered}$$

Thus, the path difference,

$$\begin{gathered} ∆\mathrm{P}={\mathrm{p}}_2-{\mathrm{P}}_1 \\ =18.0\mathrm{m}-17.0\mathrm{m} \\ =1.0\mathrm{m} \end{gathered}$$

Now,

localid="1663928116728" $$\begin{gathered} ∆\mathrm{P}={\mathrm{P}}_2-{\mathrm{P}}_1 \\ 1.0\mathrm{m}=\mathrm{°ìλ} \\ 1.0\mathrm{m}=\mathrm{k}\left(2.0\mathrm{m}\right) \\ \mathrm{k}=\frac{1}{2} \end{gathered}$$

So, the interference is destructive as the path difference is odd multiple of half wavelength with m=0.