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Step-by-Step Solution
Step 1: Recall the relationship between amplitude and intensity
The intensity $I$ of a wave is proportional to the square of its amplitude $A$. Mathematically, this can be written as
$$
I \propto A^2 \quad \Longrightarrow \quad A = \sqrt{\frac{I}{K}},
$$
where $K$ is a constant of proportionality (the same for both waves under identical medium conditions).
Step 2: Express the maximum resultant amplitude
When two waves superpose constructively, their amplitudes add directly. For amplitudes $A_1$ and $A_2$:
$$
A_{\max} = A_1 + A_2.
$$
Squaring,
$$
A_{\max}^2 = (A_1 + A_2)^2 = A_1^2 + A_2^2 + 2 A_1 A_2.
$$
Converting to intensities (using $I = K \, A^2$), we get:
$$
I_{\max} = K \, A_{\max}^2 = K \left(A_1^2 + A_2^2 + 2A_1A_2\right).
$$
Substituting $A_1 = \sqrt{\frac{I_1}{K}}$ and $A_2 = \sqrt{\frac{I_2}{K}}$:
$$
I_{\max} = K \left(\frac{I_1}{K} + \frac{I_2}{K} + 2 \sqrt{\frac{I_1}{K} \cdot \frac{I_2}{K}}\right)
= I_1 + I_2 + 2\sqrt{I_1 I_2}.
$$
Step 3: Express the minimum resultant amplitude
When two waves superpose destructively, their amplitudes subtract:
$$
A_{\min} = A_1 - A_2.
$$
Squaring,
$$
A_{\min}^2 = (A_1 - A_2)^2 = A_1^2 + A_2^2 - 2 A_1 A_2.
$$
Hence, the minimum intensity is:
$$
I_{\min} = K \, A_{\min}^2 = K \left(A_1^2 + A_2^2 - 2A_1A_2\right)
= I_1 + I_2 - 2\sqrt{I_1 I_2}.
$$
Step 4: Sum of the maximum and minimum intensities
Now, adding $I_{\max}$ and $I_{\min}$:
$$
I_{\max} + I_{\min}
= \bigl(I_1 + I_2 + 2\sqrt{I_1 I_2}\bigr) + \bigl(I_1 + I_2 - 2\sqrt{I_1 I_2}\bigr)
= 2I_1 + 2I_2.
$$
Step 5: Identify the correct answer
From the options given, $2(I_1 + I_2)$ matches our derived result. Therefore, the correct answer is:
$$
2(I_1 + I_2).
$$
