Two light beams of intensities 4,I and 9,I in | vedclass

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(A) The resultant intensity $I_R$ of two interfering beams is given by the formula: $I_R = I_1 + I_2 + 2\sqrt{I_1 I_2} \cos \phi$.At point $A$,the pha

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$24$

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(A) The resultant intensity $I_R$ of two interfering beams is given by the formula: $I_R = I_1 + I_2 + 2\sqrt{I_1 I_2} \cos \phi$.At point $A$,the phase difference $\phi = 0$. Therefore,the intensity is maximum:$I_A = I_{\max} = (\sqrt{I_1} + \sqrt{I_2})^2 = (\sqrt{9I} + \sqrt{4I})^2 = (3\sqrt{I} + 2\sqrt{I})^2 = (5\sqrt{I})^2 = 25I$.At point $B$,the phase difference $\phi = \pi$. Therefore,the intensity is minimum:$I_B = I_{\min} = (\sqrt{I_1} - \sqrt{I_2})^2 = (\sqrt{9I} - \sqrt{4I})^2 = (3\sqrt{I} - 2\sqrt{I})^2 = (\sqrt{I})^2 = I$.The difference between the resultant intensities at points $A$ and $B$ is:$I_A - I_B = 25I - I = 24I$.

Two light beams of intensities $4\,I$ and $9\,I$ interfere on a screen. The phase difference between these beams on the screen at point $A$ is $0$ and at point $B$ is $\pi$. The difference of resultant intensities,at the points $A$ and $B$,will be $....I$.

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