Two waves of the same frequency and intensities I₀ and 9I₀ produce interference. If at a certain point the resultant intensity is 7I₀, then the minimum phase difference between the two sound waves will be:
A) 90°
B) 150°
C) 120°
D) 100°

Problem Summary:
We are given two waves with the same frequency and intensities I0I_0 and 9I09I_0 producing interference. The resultant intensity at a certain point is 7I07I_0, and we need to find the minimum phase difference between the two sound waves.
Step-by-Step Solution:
1. Intensity and Interference: The intensity of a wave is proportional to the square of its amplitude. Let the amplitudes of the two waves be A1A_1 and A2A_2, where:
A12∝I0andA22∝9I0A_1^2 \propto I_0 \quad \text{and} \quad A_2^2 \propto 9I_0
Thus, the amplitudes are:
A1=I0,A2=3I0A_1 = \sqrt{I_0}, \quad A_2 = 3\sqrt{I_0}
2. Resultant Amplitude: The resultant amplitude AresA_{\text{res}} due to the interference of two waves is given by the formula:
Ares=A12+A22+2A1A2cos⁡(Δϕ)A_{\text{res}} = \sqrt{A_1^2 + A_2^2 + 2A_1A_2 \cos(\Delta \phi)}
where Δϕ\Delta \phi is the phase difference between the two waves.
The intensity of the resultant wave is proportional to the square of the resultant amplitude:
Ires∝Ares2I_{\text{res}} \propto A_{\text{res}}^2
3. Resultant Intensity Given: We are told that the resultant intensity at a certain point is 7I07I_0, so:
Ires=7I0I_{\text{res}} = 7I_0
Thus:
Ares2∝7I0A_{\text{res}}^2 \propto 7I_0
Therefore:
Ares=7⋅I0A_{\text{res}} = \sqrt{7} \cdot \sqrt{I_0}
4. Equating Intensities: From the expression for AresA_{\text{res}}, we get:
I02+(3I0)2+2(I0)(3I0)cos⁡(Δϕ)=7⋅I0\sqrt{I_0}^2 + (3\sqrt{I_0})^2 + 2(\sqrt{I_0})(3\sqrt{I_0}) \cos(\Delta \phi) = \sqrt{7} \cdot \sqrt{I_0}
Simplifying:
I0+9I0+6I0cos⁡(Δϕ)=7I0I_0 + 9I_0 + 6I_0 \cos(\Delta \phi) = 7I_0 10I0+6I0cos⁡(Δϕ)=7I010I_0 + 6I_0 \cos(\Delta \phi) = 7I_0 6I0cos⁡(Δϕ)=−3I06I_0 \cos(\Delta \phi) = -3I_0 cos⁡(Δϕ)=−12\cos(\Delta \phi) = -\frac{1}{2}
5. Finding the Phase Difference: The phase difference Δϕ\Delta \phi is given by:
Δϕ=cos⁡−1(−12)\Delta \phi = \cos^{-1}\left(-\frac{1}{2}\right)
The angle whose cosine is −12-\frac{1}{2} is 120∘120^\circ.
Final Answer:
The minimum phase difference between the two sound waves is 120°.
Correct option: C) 120°.