Two light waves of intensity I and 4I superpo | vedclass

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(C) The resultant intensity $I_R$ of two interfering waves with intensities $I_1$ and $I_2$ and phase difference $\Delta \phi$ is given by $I_R = I_1

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$4$

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(C) The resultant intensity $I_R$ of two interfering waves with intensities $I_1$ and $I_2$ and phase difference $\Delta \phi$ is given by $I_R = I_1 + I_2 + 2\sqrt{I_1 I_2} \cos(\Delta \phi)$.Given $I_1 = I$ and $I_2 = 4I$.At point $A$,$\Delta \phi = 0$:$I_{RA} = I + 4I + 2\sqrt{I \cdot 4I} \cos(0) = 5I + 2(2I)(1) = 9I$.At point $B$,$\Delta \phi = \frac{\pi}{2}$:$I_{RB} = I + 4I + 2\sqrt{I \cdot 4I} \cos(\frac{\pi}{2}) = 5I + 2(2I)(0) = 5I$.The difference in resultant intensities is $I_{RA} - I_{RB} = 9I - 5I = 4I$.

Two light waves of intensity $I$ and $4I$ superpose at point $A$ with zero phase difference and at point $B$ with a phase difference of $\frac{\pi}{2}$. Calculate the difference of resultant intensities at point $A$ and $B$. (in $I$)

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