Fraunhoffer Diffraction due to a Single Slit:
As shown in figure a source S of monochromatic light is placed at the focus of a convex lens Li. A parallel beam of light and hence a plane wavefront WW’ gets incident on a narrow rectangular slit AB of width a.
The incident wavefront disturbs all parts of the slit AB simultaneously. According to Huygens, the all parts of the slit AB will become source of secondary wavelets, which all start in the same phase. These wavelets spread out in all directions, thus causing diffraction of light after it emerges through slit AB. Suppose the diffraction pattern is focussed by a convex lens L2 on a screen placed in its focal plane.
All the secondary wavelets going straight across the slit AB are focussed at the central point O of the screen. The wavelets from any two corresponding points of the two halves of the slit reach the point O in the same phase, they add constructively to produce a central bright fringe.
Suppose, the secondary wavelets diffracted at an angle are focused at point P. The secondary wavelets start from different parts of the slit in same phase but they reach the point P in different phases. Draw perpendicular AN from A on to the ray from B. Then the path difference between the wavelets from A and B will be
BN = p = BP – AP = BM = ABsinθ = a sinθ
Let the point P be so located on the screen that the path difference p = λ and angle θ = θ1, then from the above equation, we get
a sin θ1 = λ
We can divide the slit AB into two halves AC and CB. The path difference between the wavelets from A and C will be \(\frac{\lambda}{2}\). Similarly, corresponding to every point in the upper half AC , there is a point in the lower half CB for which the path difference is \(\frac{\lambda}{2}\). Hence the wavelets from the two halves reach the point P always in opposite phases. They interfere distructively so as to produce a minimum.
Thus the condition for first dark fringe is
a sin θ1 = λ
Similarly, the condition for second dark fringe will be
a sin θ2 = 2λ
Hence the condition for nth dark fringe can be written as
a sin θn = nλ, n = 1,2,3,………….. (1)
The directions of various minima are given by
θn ≈ sinθn = \(m \frac{\lambda}{a}\)
[As λ < < d, so sinθn = θn]
Now for the positions of secondary maxima suppose the point P is so located that
p = \(\frac{3 \lambda}{2}\)
When θ = θ’1
then a sin θ’1= \(\frac{3}{2} l\)
We can divide the slit into three equal parts. The path difference between two corresponding points of the first two parts will be \(\frac{\lambda}{2}\). The wavelets from these points will interfere destructively. However, the wavelets from the third part of the slit will contribute to some intensity forming a secondary maximum. The intensity of this maximum is much less than that of the central maximum.
Thus the condition for the first secondary maximum is
a sin θ’1 = \(\frac{3}{2} \lambda\)
Similarly, the condition for the second secondary maximum is
a sin θ’2 = \(\frac{5}{2} \lambda\)
Hence the condition for nthsecondary maximum can be written as
d sin θ’n = (2n + 1)\(\frac{\lambda}{2}\) , n = 1, 2, 3,…………. (2)
The directions of secondary maxima are given by
θ’n = sin θ’n = (2n + 1)\(\frac{\lambda}{2d}\), n = 1, 2, 3, ………….. (3)
The intensity of secondary maxima decreases as n increases.
