Two beams of light having intensities I and 4 | vedclass

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(B) The resultant intensity $I_{net}$ for two interfering beams is given by the formula: $I_{net} = I_1 + I_2 + 2\sqrt{I_1 I_2} \cos(\Delta \phi)$.At

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$4$

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(B) The resultant intensity $I_{net}$ for two interfering beams is given by the formula: $I_{net} = I_1 + I_2 + 2\sqrt{I_1 I_2} \cos(\Delta \phi)$.At point $A$,the phase difference $\Delta \phi = \pi/2$. Substituting the values $I_1 = I$ and $I_2 = 4I$:$I_A = I + 4I + 2\sqrt{I \cdot 4I} \cos(\pi/2) = 5I + 2(2I)(0) = 5I$.At point $B$,the phase difference $\Delta \phi = \pi$. Substituting the values:$I_B = I + 4I + 2\sqrt{I \cdot 4I} \cos(\pi) = 5I + 2(2I)(-1) = 5I - 4I = I$.The difference between the resultant intensities at $A$ and $B$ is:$|I_A - I_B| = |5I - I| = 4I$.

Two beams of light having intensities $I$ and $4I$ interfere to produce a fringe pattern on a screen. The phase difference between the beams 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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