A parallel beam of monochromatic light is used in a Young's double slit experiment. The slits are separated by a distance d and the screen is placed parallel to the plane of the slits. Show that if the incident beam makes an angle `theta=sin^-1(lamda/(2d))` with the normal to the plane of the slits, there will be a dark fringe at the centre Po of the pattern.

Answer

maximum

Explanation

clearly, at `O`, no defference of path due to inside motion of rays.
only path difference `= S_(1) Q`
`= d cos (90-theta)`
`= d sin theta`.
`= dxx(lambda)/(2d)=(lambda)/(2)`
Net phase difference
so `Deltaphi=(lambda)/(2)xx(2pi)/(lambda)+-(mu-1)t.(2pi)/(lambda)=pi+-(mu-1)(lamda)/(2(mu-1))(2pi)/(lambda)=2pi,0`
`rArr` constructive interference and hence `I_(res)=maxima`