Two light beams having intensities I and 9I i | vedclass

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(B) 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

$6$

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(B) The resultant intensity $I_R$ for two interfering beams with intensities $I_1$ and $I_2$ and phase difference $\phi$ is given by $I_R = I_1 + I_2 + 2\sqrt{I_1 I_2} \cos \phi$.At point $P$,the phase difference is $\phi_P = \frac{\pi}{2}$.$I_P = I + 9I + 2\sqrt{I 	imes 9I} \cos \frac{\pi}{2} = 10I + 2(3I)(0) = 10I$.At point $Q$,the phase difference is $\phi_Q = \pi$.$I_Q = I + 9I + 2\sqrt{I 	imes 9I} \cos \pi = 10I + 2(3I)(-1) = 10I - 6I = 4I$.The difference between the resultant intensities at $P$ and $Q$ is:$I_P - I_Q = 10I - 4I = 6I$.

Two light beams having intensities $I$ and $9I$ interfere to produce a fringe pattern on a screen. The phase difference between the beams is $\frac{\pi}{2}$ at point $P$ and $\pi$ at point $Q$. The difference between the resultant intensities at $P$ and $Q$ will be ....... $I$.

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