Young’s double slit experiment to determine the fringe width.
Consider ‘s’ be the point source, which emits the monochromatic light of wave lengths let S1 and S2 be the coherent sources emitted from single source (point) ‘s’ which are separated by distance ‘d’.
Let screen is placed at distance ‘s’ from the slit as in the figure. At any instant the wave emitting from S1 and S2 will meet at point ‘p’ which is at the distance y from the center of the screen O.
Let x be path difference of wave emitting from S1 and S2 and meeting at ‘p’.
The point ‘p’ will be the position of minimum intensity, if x = nλ
The difference of distance of conservative bright fringes from the center of fringe gives the width of dark fringe.
The point p will be the position of minimum intensity, if
The difference of distance of conservative dark fringe from center of fringe gives width of bright fringe.
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