Two sound waves having intensities I and 4I i | vedclass

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(B) The resultant intensity $I_R$ of two interfering waves is given by the formula: $I_R = I_1 + I_2 + 2\sqrt{I_1 I_2} \cos \phi$,where $\phi$ is the

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(B) The resultant intensity $I_R$ of two interfering waves is given by the formula: $I_R = I_1 + I_2 + 2\sqrt{I_1 I_2} \cos \phi$,where $\phi$ is the phase difference.Given $I_1 = I$ and $I_2 = 4I$.At point $A$,the phase difference $\phi_A = \pi / 2$. Substituting this into the formula:$I_A = I + 4I + 2\sqrt{I \cdot 4I} \cos(\pi / 2) = 5I + 2(2I)(0) = 5I$.At point $B$,the phase difference $\phi_B = \pi$. Substituting this into the formula:$I_B = I + 4I + 2\sqrt{I \cdot 4I} \cos(\pi) = 5I + 2(2I)(-1) = 5I - 4I = I$.The difference between the resultant intensities at $A$ and $B$ is $I_A - I_B = 5I - I = 4I$.

Two sound waves having intensities $I$ and $4I$ interfere to produce an interference pattern. The phase difference between the waves is $\pi / 2$ at point $A$ and $\pi$ at point $B$. Then the difference between the resultant intensities at $A$ and $B$ is (in $I$)

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