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(A) At the centre of a bright fringe $(P_1)$,the intensity is maximum,$I_1 = I_{max} = 4I_0$,where $I_0$ is the intensity of each individual slit.The

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$2$

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(A) At the centre of a bright fringe $(P_1)$,the intensity is maximum,$I_1 = I_{max} = 4I_0$,where $I_0$ is the intensity of each individual slit.The path difference $\Delta x$ at a distance $y$ from the centre is given by $\Delta x = \frac{yd}{D}$.Given $y = \frac{\beta}{4}$,where $\beta = \frac{\lambda D}{d}$ is the fringe width.Thus,$\Delta x = \frac{(\lambda D / 4d) \cdot d}{D} = \frac{\lambda}{4}$.The phase difference $\phi$ is given by $\phi = \frac{2\pi}{\lambda} \cdot \Delta x = \frac{2\pi}{\lambda} \cdot \frac{\lambda}{4} = \frac{\pi}{2}$.The intensity at any point is given by $I = I_{max} \cos^2(\frac{\phi}{2})$.For point $P_2$,$I_2 = I_1 \cos^2(\frac{\pi/2}{2}) = I_1 \cos^2(\frac{\pi}{4})$.Since $\cos(\frac{\pi}{4}) = \frac{1}{\sqrt{2}}$,we have $\cos^2(\frac{\pi}{4}) = \frac{1}{2}$.Therefore,$I_2 = I_1 \cdot \frac{1}{2}$,which implies $\frac{I_1}{I_2} = 2$.

In the Young's double slit experiment,the intensities at two points $P_1$ and $P_2$ on the screen are respectively $I_1$ and $I_2$. If $P_1$ is located at the centre of a bright fringe and $P_2$ is located at a distance equal to a quarter of fringe width from $P_1$,then $\frac{I_1}{I_2}$ is

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