In a Young's double slit experiment,{I_o} is | vedclass

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(C) The path difference at point $P$ is given by $\Delta = \frac{xd}{D}$.Since the fringe width $\beta = \frac{\lambda D}{d}$,we can write the path di

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

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(C) The path difference at point $P$ is given by $\Delta = \frac{xd}{D}$.Since the fringe width $\beta = \frac{\lambda D}{d}$,we can write the path difference as $\Delta = \frac{x\lambda}{\beta}$.The phase difference $\phi$ is given by $\phi = \frac{2\pi}{\lambda} \Delta = \frac{2\pi}{\lambda} \cdot \frac{x\lambda}{\beta} = \frac{2\pi x}{\beta}$.The intensity at any point is given by $I = I_{max} \cos^2(\frac{\phi}{2})$.Given $I_{max} = I_o$,the intensity at point $P$ is $I = I_o \cos^2(\frac{2\pi x}{2\beta}) = I_o \cos^2(\frac{\pi x}{\beta})$.

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

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