Two coherent narrow slits emitting sound of wavelength λ in the...

The path difference at point $O$ is given by:

$\Delta x_O = S_1O - S_2O = D - (D - 2\lambda) = 2\lambda$

Since the path difference is an integer multiple of the wavelength, constructive interference occurs at $O$, resulting in maximum intensity.

For the intensity at point $P$ to match the intensity at $O$, constructive interference must also occur at $P$. The path difference at $P$ is:

$\Delta x_P = S_1P - S_2P = \sqrt{D^2 + x^2} - \sqrt{(D - 2\lambda)^2 + x^2}$

As $x$ increases, the path difference decreases from $2\lambda$ towards $0$. The next maximum intensity occurs when the path difference is $\lambda$. Thus, we set:

$\sqrt{D^2 + x^2} - \sqrt{(D - 2\lambda)^2 + x^2} = \lambda$

Let $y = \sqrt{D^2 + x^2}$. The equation becomes:

$y - \sqrt{y^2 - 4D\lambda + 4\lambda^2} = \lambda$

$y - \lambda = \sqrt{y^2 - 4D\lambda + 4\lambda^2}$

Squaring both sides, we get:

$y^2 - 2y\lambda + \lambda^2 = y^2 - 4D\lambda + 4\lambda^2$

$2y\lambda = 4D\lambda - 3\lambda^2$

$y = 2D - 1.5\lambda$

Given that $D \gg \lambda$, we can approximate $y \approx 2D$. Substituting back $y = \sqrt{D^2 + x^2}$:

$\sqrt{D^2 + x^2} = 2D$

Squaring both sides:

$D^2 + x^2 = 4D^2$

$x^2 = 3D^2$

$x = \sqrt{3}D$

Answer: $\sqrt{3} D$