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(B) The intensity at any point in an interference pattern is given by $I = I_0 \cos^2 \left( \frac{\phi}{2} \right)$,where $\phi$ is the phase differe

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$\frac{4}{3}$

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(B) The intensity at any point in an interference pattern is given by $I = I_0 \cos^2 \left( \frac{\phi}{2} ight)$,where $\phi$ is the phase difference.The relationship between phase difference $\phi$ and path difference $\Delta x$ is $\phi = \frac{2\pi}{\lambda} \Delta x$.Given the path difference $\Delta x = \frac{\lambda}{6}$,the phase difference is $\phi = \frac{2\pi}{\lambda} 	imes \frac{\lambda}{6} = \frac{\pi}{3}$.Substituting this into the intensity formula:$I = I_0 \cos^2 \left( \frac{\pi/3}{2} ight) = I_0 \cos^2 \left( \frac{\pi}{6} ight)$.Since $\cos \left( \frac{\pi}{6} ight) = \frac{\sqrt{3}}{2}$,we have:$I = I_0 \left( \frac{\sqrt{3}}{2} ight)^2 = I_0 	imes \frac{3}{4}$.Therefore,the ratio $\frac{I_0}{I} = \frac{4}{3}$.

In a Young's double slit experiment,the intensity at a point where the path difference is $\frac{\lambda}{6}$ ($\lambda$ being the wavelength of the light used) is $I$. If $I_0$ denotes the maximum intensity,the ratio $\frac{I_0}{I}$ is equal to:

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