(a) Diffraction of light at a single slit : When monochromatic light is made incident on a single slit, we get diffraction pattern on a screen placed behind the slit. The diffraction pattern contains bright and dark bands, the intensity of central band is maximum and goes on decreasing on both sides.
Explanation : Let AB be a slit of width ‘a’ and a parallel beam of monochromatic light is incident on it. According to Fresnel the diffraction pattern is the result of superposition of a large number of waves, starting from different points of illuminated slit.
Let θ be the angle of diffraction for waves reaching at point P of screen and AN the perpendicular dropped from A on wave diffracted from B.
The path difference between rays diffracted at points A and B,
Δ = BP - AP = BN
To find the effect of all coherent waves at P, we have to sum up their contribution, each with a different phase. This was done by Fresnel by rigorous calculations, but the main features may be explained by simple arguments given below :
At the central point C of the screen, the angle θ is zero. Hence the waves starting from all points of slit arrive in the same phase. This gives maximum intensity at the central point C. If point P on screen is such that the path difference between rays starting from edges A and B is λ, then path difference
Minima: Now we divide the slit into two equal halves AO and OB, each of width a/2. Now for every point, M1 in AO, there is a corresponding point M2 in OB, such that M1 M2 = a/2; Then path difference between waves arriving at P and starting from M1 and M2 will be a/2sinθ = λ/2. This means that the contributions from the two halves of slit AO and OB are opposite in phase and so cancel each other. Thus equation (2) gives the angle of diffraction at which intensity falls to zero. Similarly it may be shown that the intensity is zero for sinθ = nλ/a, with n as integer. Thus the general condition of minima is
asinθ = nλ...(iii)
Secondary Maxima: Let us now consider angle θ such that
which is midway between two dark bands given by
Let us now divide the slit into three parts. If we take the first two of parts of slit, the path difference between rays diffracted from the extreme ends of the first two parts
Then the first two parts will have a path difference of λ/2 and cancel the effect of each other. The remaining third part will contribute to the intensity at a point between two minima. Clearly there will be a maxima between first two minima, but this maxima will be of much weaker intensity than central maximum. This is called first secondary maxima. In a similar manner we can show that there are secondary maxima between any two consecutive minima; and the intensity of maxima will go on decreasing with increase of order of maxima. In general the position of nth maxima will be given by
a sinθ = (n + 1/2)λ [n =1, 2, 3, 4, .... ] ...(iv)
The intensity of secondary maxima decrease with increase of order n because with increasing n, the contribution of slit decreases.
For n = 2, it is one-fifth, for n = 3, it is one-seventh and so on.
(b) Width of central Maxima ' β' = 2Dλ/a
a → size of slit
If size of slit is doubled, width of central maxima becomes half. Intensity varies as square of slit width. It width of slit is doubled, intensity gets four times.
