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# In Young's double slit experiment the distance between adjacent fringe is x and L is distance between slit and screen. If d is split separation , then wave length is

$(a)\;\frac{xd}{L} \\ (b)\;\frac{xL}{d} \\ (c)\;\frac{2Ld}{x} \\ (d)\;\frac{1}{Lxd}$

Fringe width $\beta =\large\frac{D \lambda}{d}$
Here fringe width =x , and distance between screen and slit = L
$\therefore x=\large\frac{L \lambda}{d}$
$\lambda =\large\frac{xd}{L}$
Hence a is the correct answer.

edited Jul 17, 2014