Interference fringes are produced on a screen | vedclass

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

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$6$

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(C) The resultant intensity $I_R$ for two interfering waves 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 $I_2 = 9I$.At point $P$,the phase difference $\phi_P = \frac{\pi}{2}$.$I_P = I + 9I + 2\sqrt{I \cdot 9I} \cos(\frac{\pi}{2}) = 10I + 6I(0) = 10I$.At point $Q$,the phase difference $\phi_Q = \pi$.$I_Q = I + 9I + 2\sqrt{I \cdot 9I} \cos(\pi) = 10I + 6I(-1) = 10I - 6I = 4I$.The difference between the resultant intensities at points $P$ and $Q$ is $\Delta I = I_P - I_Q = 10I - 4I = 6I$.

Interference fringes are produced on a screen by using two light sources of intensities $I$ and $9I$. The phase difference between the beams is $\frac{\pi}{2}$ at the point $P$ and $\pi$ at the point $Q$ on the screen. The difference between the resultant intensities at point $P$ and $Q$ is (in $I$)

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