Two sound sources, S_1 and S_2 , emitting waves of identical...

Given wavelength $\lambda = 20.0 \text{ cm}$, distance between sources $d = 20.0 \text{ cm}$, and distance to the line of movement $D = 20.0 \text{ cm}$.

For the first minimum, the path difference must be $\Delta x = \dfrac{\lambda}{2} = 10.0 \text{ cm}$.

Let the sources be located at $(-10, 0)$ and $(10, 0)$. The detector is initially at $(0, 20)$ and moves to a new position $(x, 20)$.

The distances from the sources to the detector are $S_1 P = \sqrt{(x + 10)^2 + 20^2}$ and $S_2 P = \sqrt{(x - 10)^2 + 20^2}$.

The path difference is $\Delta x = S_1 P - S_2 P = 10$.

$\sqrt{(x + 10)^2 + 400} - \sqrt{(x - 10)^2 + 400} = 10$

$\sqrt{(x + 10)^2 + 400} = 10 + \sqrt{(x - 10)^2 + 400}$

Squaring both sides:

$(x + 10)^2 + 400 = 100 + (x - 10)^2 + 400 + 20 \sqrt{(x - 10)^2 + 400}$

$x^2 + 20x + 500 = x^2 - 20x + 600 + 20 \sqrt{(x - 10)^2 + 400}$

$40x - 100 = 20 \sqrt{(x - 10)^2 + 400}$

Dividing by 20:

$2x - 5 = \sqrt{(x - 10)^2 + 400}$

Squaring both sides again:

$(2x - 5)^2 = (x - 10)^2 + 400$

$4x^2 - 20x + 25 = x^2 - 20x + 100 + 400$

$3x^2 = 475$

$x = \sqrt{\dfrac{475}{3}} \approx \sqrt{158.33} \approx 12.58 \text{ cm}$

Rounding to one decimal place, the minimum distance the detector must be shifted is $12.6 \text{ cm}$.