In the Young's double slit experiment,the int | vedclass

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(A) Let $I_{max}$ be the intensity at the center of the bright fringe $(P_1)$. Thus,$I_1 = I_{max}$.The path difference $\Delta x$ corresponding to a

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$2$

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(A) Let $I_{max}$ be the intensity at the center of the bright fringe $(P_1)$. Thus,$I_1 = I_{max}$.The path difference $\Delta x$ corresponding to a distance $y$ from the center is given by $\Delta x = \frac{yd}{D}$.Given $y = \frac{\beta}{4}$,where $\beta = \frac{\lambda D}{d}$ is the fringe width.Substituting $y$,we get $\Delta x = \frac{(\lambda D / 4d) \cdot d}{D} = \frac{\lambda}{4}$.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 $I = I_{max} \cos^2(\phi/2)$.For point $P_2$,$I_2 = I_{max} \cos^2(\pi/4) = I_{max} \cdot (1/\sqrt{2})^2 = \frac{I_{max}}{2}$.Therefore,the ratio $I_1/I_2 = I_{max} / (I_{max}/2) = 2$.

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 $I_1/I_2$ is:

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