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(D) The phase difference $\Delta \phi$ is related to the path difference $\Delta x$ by the formula $\Delta \phi = \frac{2 \pi}{\lambda} \Delta x$.Give

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$\frac{3}{4}$

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(D) The phase difference $\Delta \phi$ is related to the path difference $\Delta x$ by the formula $\Delta \phi = \frac{2 \pi}{\lambda} \Delta x$.Given $\Delta x = \frac{\lambda}{6}$,we have $\Delta \phi = \frac{2 \pi}{\lambda} \cdot \frac{\lambda}{6} = \frac{\pi}{3}$.The intensity $I$ at any point is given by $I = I_0 \cos^2 \left( \frac{\Delta \phi}{2} \right)$,where $I_0$ is the maximum intensity.Substituting the value of $\Delta \phi$,we get $I = I_0 \cos^2 \left( \frac{\pi/3}{2} \right) = I_0 \cos^2 \left( \frac{\pi}{6} \right)$.Since $\cos \left( \frac{\pi}{6} \right) = \frac{\sqrt{3}}{2}$,we have $I = I_0 \left( \frac{\sqrt{3}}{2} \right)^2 = I_0 \cdot \frac{3}{4}$.Therefore,$\frac{I}{I_0} = \frac{3}{4}$.

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,$I/I_0$ is equal to

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