Step 1: Understanding the Concept:
Beats are produced by the superposition of two waves with nearly equal frequencies. The intensity of a wave is proportional to the square of its amplitude. At the point of maximum intensity (constructive interference), the amplitudes of the two waves add up.
Step 2: Key Formula or Approach:
Let the amplitude of each source be $A_0$. The intensity of a single source is $I_0 \propto A_0^2$.
When the waves interfere constructively, the resultant amplitude $A_{max}$ is the sum of the individual amplitudes.
\[ A_{max} = A_0 + A_0 = 2A_0 \]
The maximum intensity, $I_{max}$, is proportional to the square of the resultant maximum amplitude.
\[ I_{max} \propto A_{max}^2 \]
We need to compare $I_{max}$ with $I_0$.
Step 3: Detailed Explanation:
Let the intensity of one source be $I_0$. We have the relationship:
\[ I_0 = k A_0^2 \]
where k is a constant of proportionality.
At the location of a beat maximum, the two waves are in phase, and their amplitudes add up. The resultant amplitude is:
\[ A_{max} = A_0 + A_0 = 2A_0 \]
The maximum intensity is then:
\[ I_{max} = k (A_{max})^2 = k (2A_0)^2 = k (4A_0^2) \]
Now, substitute $I_0 = k A_0^2$ into this equation:
\[ I_{max} = 4 (k A_0^2) = 4 I_0 \]
Thus, the maximum intensity of the beats is four times the intensity of a single source.
Step 4: Final Answer:
The maximum intensity of beats will be four times when compared to that of one source. Therefore, option (C) is correct.