In an interference experiment,the path differ | vedclass

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(C) The intensity $I$ at any point is given by $I = I_{max} \cos^2(\phi/2)$,where $\phi$ is the phase difference and $\phi = (2\pi/\lambda) \Delta x$.

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$1$/$3$

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(C) The intensity $I$ at any point is given by $I = I_{max} \cos^2(\phi/2)$,where $\phi$ is the phase difference and $\phi = (2\pi/\lambda) \Delta x$.For point $A$: $\Delta x_A = \lambda/3 \implies \phi_A = (2\pi/\lambda)(\lambda/3) = 2\pi/3$.$I_A = I_{max} \cos^2(\phi_A/2) = I_{max} \cos^2(\pi/3) = I_{max} (1/2)^2 = I_{max}/4$.For point $B$: $\Delta x_B = \lambda/6 \implies \phi_B = (2\pi/\lambda)(\lambda/6) = \pi/3$.$I_B = I_{max} \cos^2(\phi_B/2) = I_{max} \cos^2(\pi/6) = I_{max} (\sqrt{3}/2)^2 = 3I_{max}/4$.Ratio $I_A / I_B = (I_{max}/4) / (3I_{max}/4) = 1/3$.

In an interference experiment,the path difference between two interfering waves at a point $A$ on the screen is $\lambda/3$,where $\lambda$ is the wavelength of these waves,and at another point $B$ the path difference is $\lambda/6$. The ratio of intensities at points $A$ and $B$ is . . . . . . .

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