In a Young's double slit experiment,I_0 is th | vedclass

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(C) The path difference at point $P$ is given by $\Delta x = \frac{xd}{D}$.The phase difference $\phi$ at point $P$ is $\phi = \frac{2\pi}{\lambda} \D

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$I_0 \cos^2 \frac{\pi x}{\beta}$

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(C) The path difference at point $P$ is given by $\Delta x = \frac{xd}{D}$.The phase difference $\phi$ at point $P$ is $\phi = \frac{2\pi}{\lambda} \Delta x = \frac{2\pi}{\lambda} \frac{xd}{D}$.Since the fringe width $\beta = \frac{\lambda D}{d}$,we can write $\phi = \frac{2\pi x}{\beta}$.The intensity at any point is given by $I = I_1 + I_2 + 2\sqrt{I_1 I_2} \cos \phi$. Assuming $I_1 = I_2 = I$,we have $I = 2I + 2I \cos \phi = 4I \cos^2 \frac{\phi}{2}$.Given the intensity at the central maximum is $I_0 = 4I$,we substitute $I = \frac{I_0}{4}$ and $\phi = \frac{2\pi x}{\beta}$:$I_P = 4 \left( \frac{I_0}{4} \right) \cos^2 \left( \frac{2\pi x / \beta}{2} \right) = I_0 \cos^2 \frac{\pi x}{\beta}$.

In a Young's double slit experiment,$I_0$ is the intensity at the central maximum and $\beta$ is the fringe width. The intensity at a point $P$ at a distance $x$ from the central maximum will be

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