CBSE Class 12-science Answered
Diffraction of Light at a Single Slit
Consider a parallel beam of light from a lens falling on a slit AB. As diffraction occurs, the pattern is focused on the screen XY with the help of lens L2.
Each point on the unblocked portion of plane wavefront AB sends out secondary wavelets in all directions. The waves from points equidistant from the center C lying on the upper and lower half reach the point O with path difference being zero and hence reinforce each other producing a maximum intensity at the point 0.
Consider a point 'P' on the screen. The intensity at P will depend on the path difference between the secondary waves emitted from the corresponding point of the wavefront. Consider a set of waves making an angle q with CO. Draw AN perpendicular to BK. The path difference between the secondary waves reaching P from
A and B = BN = AB Sin q = a sin q
where AB = a width of the aperture.
If BN = a (i.e., path difference) the whole wave front can be considered to be divided into two halves, i.e., CA and CB. The path difference between A and C will be l/2 and so for every point in the upper half AC, there is a corresponding point in the lower half CB for which the path difference between the waves is l/2. So P is a point of destructive interference.
qn is the angle of diffraction of nth dark fringe from O.
Hope this clarifies your doubt.
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