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In the Young's double slit experiment, the resultant intensity at a point on the screen is $75 \%$ of the maximum intensity of the bright fringe. Then the phase difference between the two interfering rays at that point is :

\[\begin {array} {1 1} (a)\;\frac{\pi}{6} & \quad (b)\;\frac{\pi}{4} \\ (c)\;\frac{\pi}{3} & \quad (d)\;\frac{\pi}{2} \end {array}\]

1 Answer

answered Nov 7, 2013 by pady_1

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