Step-by-step Derivation

1. Relating Intensity and Amplitude: The intensity $I$ of a wave is proportional to the square of its amplitude $A$. Let the proportionality constant be $k$. Since all four waves have the same intensity $I_0$, they must also have the same amplitude, which we'll call $A_0$. $I_0 = k A_0^2$

2. Using the Phasor Method for Superposition: When waves superpose, the resultant amplitude can be found by vectorially adding the amplitudes of the individual waves, considering their phase differences. This can be visualized using phasors. Each wave is represented by a phasor of length $A_0$ at an angle equal to its phase. The four given phase angles are $\phi_1 = 0$, $\phi_2 = \frac{\pi}{3}$, $\phi_3 = \frac{2\pi}{3}$, and $\phi_4 = \pi$.

3. Calculating Components of the Resultant Amplitude: We resolve each phasor into its x and y components and sum them up to find the components of the resultant phasor, $A_x$ and $A_y$.

   • Sum of x-components ($A_x$): $A_x = A_0 \cos(0) + A_0 \cos\left(\frac{\pi}{3}\right) + A_0 \cos\left(\frac{2\pi}{3}\right) + A_0 \cos(\pi)$ We know the values of the cosine functions: $\cos(0) = 1$ $\cos(\frac{\pi}{3}) = \frac{1}{2}$ $\cos(\frac{2\pi}{3}) = -\frac{1}{2}$ $\cos(\pi) = -1$ Substituting these values: $A_x = A_0 \left(1 + \frac{1}{2} - \frac{1}{2} - 1\right) = A_0(0) = 0$

   • Sum of y-components ($A_y$): $A_y = A_0 \sin(0) + A_0 \sin\left(\frac{\pi}{3}\right) + A_0 \sin\left(\frac{2\pi}{3}\right) + A_0 \sin(\pi)$ We know the values of the sine functions: $\sin(0) = 0$ $\sin(\frac{\pi}{3}) = \frac{\sqrt{3}}{2}$ $\sin(\frac{2\pi}{3}) = \frac{\sqrt{3}}{2}$ $\sin(\pi) = 0$ Substituting these values: $A_y = A_0 \left(0 + \frac{\sqrt{3}}{2} + \frac{\sqrt{3}}{2} + 0\right) = A_0 \left(2 \cdot \frac{\sqrt{3}}{2}\right) = A_0 \sqrt{3}$

4. Calculating the Resultant Amplitude ($A_{res}$): The magnitude of the resultant amplitude is found using the Pythagorean theorem: $A_{res} = \sqrt{A_x^2 + A_y^2}$ $A_{res} = \sqrt{0^2 + (A_0 \sqrt{3})^2} = \sqrt{3A_0^2} = A_0 \sqrt{3}$

5. Calculating the Resultant Intensity ($I_{res}$): The resultant intensity is proportional to the square of the resultant amplitude: $I_{res} = k A_{res}^2$ $I_{res} = k (A_0 \sqrt{3})^2 = k (3A_0^2) = 3 (k A_0^2)$

6. Finding the value of n: We know from step 1 that $I_0 = k A_0^2$. Substituting this into the expression for $I_{res}$: $I_{res} = 3 I_0$ The problem states that the intensity of the resulting wave is $nI_0$. Comparing this with our result: $nI_0 = 3I_0$ $n = 3$

Conclusion

The value of n is 3.