Two beams of light having intensities I and 4 | vedclass

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(A) The resultant intensity $I_R$ for two interfering beams with intensities $I_1$ and $I_2$ and phase difference $\phi$ is given by $I_R = I_1 + I_2

Vedclass · Accepted Answer

$4$

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(A) The resultant intensity $I_R$ for two interfering beams with intensities $I_1$ and $I_2$ and phase difference $\phi$ is given by $I_R = I_1 + I_2 + 2\sqrt{I_1 I_2} \cos \phi$.Given $I_1 = I$ and $I_2 = 4I$.At point $A$,the phase difference $\phi_A = \pi / 2$. Thus,$I_A = I + 4I + 2\sqrt{I \cdot 4I} \cos(\pi / 2) = 5I + 4I(0) = 5I$.At point $B$,the phase difference $\phi_B = \pi$. Thus,$I_B = I + 4I + 2\sqrt{I \cdot 4I} \cos(\pi) = 5I + 4I(-1) = 5I - 4I = I$.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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