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(A) From the figure,the distance $S_1D = 8\,m$ and the distance $S_1S_2 = 6\,m$. Since $S_1S_2D$ forms a right-angled triangle,the distance $S_2D = \s

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$4\,I_0$

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(A) From the figure,the distance $S_1D = 8\,m$ and the distance $S_1S_2 = 6\,m$. Since $S_1S_2D$ forms a right-angled triangle,the distance $S_2D = \sqrt{(S_1D)^2 + (S_1S_2)^2} = \sqrt{8^2 + 6^2} = \sqrt{64 + 36} = 10\,m$.The path difference is $\Delta x = S_2D - S_1D = 10\,m - 8\,m = 2\,m$.The phase difference $\Delta \phi$ is given by $\Delta \phi = \frac{2\pi}{\lambda} \Delta x = \frac{2\pi}{8} 	imes 2 = \frac{\pi}{2}$.The resultant intensity $I$ for two sources of equal intensity $I_1 = I_2 = 2I_0$ is given by $I = I_1 + I_2 + 2\sqrt{I_1 I_2} \cos(\Delta \phi)$.Substituting the values: $I = 2I_0 + 2I_0 + 2\sqrt{(2I_0)(2I_0)} \cos(\frac{\pi}{2})$.Since $\cos(\frac{\pi}{2}) = 0$,we get $I = 2I_0 + 2I_0 + 0 = 4I_0$.

In the figure,the intensity of waves arriving at $D$ from two coherent sources $S_1$ and $S_2$ is $2I_0$ each. The wavelength of the wave is $\lambda = 8\,m$. The resultant intensity at $D$ will be:

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