91Ó°ÊÓ

Wavelength of sound$\mathrm{λ}=40.00\text{cm}=40×{10}^{−2}\text{m}$.

The difference between the distance traveled by the sound wave, when it goes around a half circle and across the diameter of the circle, should satisfy the condition for destructive interference. Using this concept, find the radius of the circle.

Formula is as follow:

$\mathrm{f}=\frac{2\mathrm{π}}{\mathrm{λ}}△\mathrm{L}$

where, L is length, $\mathrm{λ}$ is wavelength and f is frequency.

Phase difference between two sound waves can be calculated by equation,

$\mathrm{ϕ}=\frac{2\mathrm{π}}{\mathrm{λ}}\text{Δ}\mathrm{L}$

Here, $\text{Δ}\mathrm{L}$is the path difference between two sound waves,

$\text{Δ}\mathrm{L}=\mathrm{Ï€°ù}−2\mathrm{r}$

role="math" localid="1661493583341" $\mathrm{Ï•}=\frac{2\mathrm{Ï€}}{\mathrm{λ}}(\mathrm{Ï€°ù}−2\mathrm{r})$

But phase difference for destructive interference $\mathrm{Ï•}=\mathrm{Ï€}$which is the smallest value for destructive interference is,

$\mathrm{Ï€}=\frac{2\mathrm{Ï€}}{\mathrm{λ}}(\mathrm{Ï€°ù}−2\mathrm{r})$

$1=\frac{2}{\mathrm{λ}}(\mathrm{π}−2)\mathrm{r}$

$\mathrm{r}=\frac{\mathrm{λ}}{2(\mathrm{π}−2)}$

${\mathrm{r}}_{\min }=\frac{40×{10}^{−2}\text{″¾}}{2\left(1.14\right)}$

${\mathrm{r}}_{\min }=\frac{40×{10}^{−2}\text{″¾}}{2.28}$

${\mathrm{r}}_{\min }=17.5\text{cm}$

Hence, the smallest radius$\mathrm{r}$ that results in minimum intensity at the detector is ${\mathrm{r}}_{\min }=17.5\text{cm}$.