how to find the position of the second dark fringe in a diffraction pattern of a monochromatic light through a single slit?

 

dear student,

to locate the dark fringes, we shall use a clever strategy that involves pairing up all the rays coming through the slit and finding what conditions cause the waves of the rays in each pair to cancel each other.  first, we divide the slit into two zones of equal widths a/2.  then we extend a light ray from the top point of the top zone, and another from the top point of the bottom zone, to the same point on the screen.

 

 

 

 

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figure 1
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            the waves of the pair of rays are in phase within the slit because they originate from the same wavefront passing through the slit.  however, tto produce he first dark fringe they must be out of phase by l/2 when they reach the viewing screen.  this phase difference is due to their path length difference.  to display this path length difference, we find a point b on the second ray such that the distance from that point to the screen is the same as the path length for the first ray.  then the phase difference is the distance from b to the center of the slit.  the two rays are approximately parallel if the distance to the screen is much greater than the slit with, a.  this makes the triangle formed by the top of the slit, the center of the slit and point b a right triangle.  therefore, using a little geometry, we find that one of the angles of this triangle is q.  the path length difference between the two rays is then equal to (a/2)sinq.

            this will be true for any pair of rays originating from corresponding points in these two regions going to the same point.  setting this common path length difference equal to l/2, we have

 

a/2 sinq = l/2,

giving us

                                                a sinq = l (first minimum).                                               (1)

 

given the slit width a and the wavelength l, this tells us the angle q of the first dark fringe above and (by symmetry) below the central axis.

            we can find the second dark fringes above and below the central axis in the same way, except that we now divide the slit into *four* zones of equal widths a/4. and extend these four rays to the same point on the screen.  to produce a dark fringe, the path length difference between adjacent pairs of rays must be l/2.  so we now have

 

a/4 sinq = l/2,

giving us

                                                a sinq = 2l (second minimum)                                         (2)

 

            we can continue to locate dark fringes in the diffraction pattern by splitting up the slit into more zones of equal width.  we would always choose an even number of zones so that the waves could be paired as we have been doing.  in this way we would find that the dark fringes can be located with the following general equation:

 

                        a sinq = ml  for m = 1, 2, 3, . . . (minima – dark fringes).  
 
thankyou
team
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