Fraunhofer diffraction at a single slit:
Suppose a parallel beam of light incident normally on a slit AB of width d after refraction through the lens L1. These rays diffract and after diffraction, the beam is focused on the screen XY by means of lens L2.
Result:
A wide control central bright fringe of maximum intensity is obtained at the center P of the screen and on either side of the central fringe, dark and bright fringes of decreasing intensity are observed. These bends are referred to as secondary minima and maxima respectively. It is found that:
- The width of central maxima is double that of secondary maxima and
- The intensity of the secondary maxima goes on decreasing.
Explanation: According to Huygen's hypothesis, when light falls on the slit, it becomes a source of secondary wavelets. These wavelets are initially in phase and spread out in all directions.
- Central Maxima: Consider a point P at the center of the screen as shown in the figure. Light travels equal optical paths to reach point P and are in the same phase. Thus, P is the position of maximum intensity which is known as the central (or principal) maxima.
- Position of the secondary maxima and minima: Suppose a point Q on the screen at which wavelets traveling in a direction making an angle θ with CP are brought to focus by the lens. The wavelets from different parts of the slit will not reach point Q in phase, although, they are initially in phase. this is because they cover unequal distances in reaching point Q. The wavelets from points A and B will have a path difference BN.
i.e., BQ - AQ = BN = d sin (θ)
Or, \(\lambda=d\sin\left(\theta\right)\) \(\left[\because\sin\left(\theta\right)=\frac{BN}d\right]\)
- For nth secondary minima, we have,
Path difference = nλ ( n = 1, 2, 3, ...)
\(\Rightarrow d\sin\left(\theta_n\right)=n\lambda\)
Or, \(\sin\left(\theta_n\right)=\frac{n\lambda}d\)
Since θ is small, sin(θ) ≅ θn
∴ \(\theta_n=\frac{n\lambda}d\)
For the first minima, n = 1 \(\Rightarrow\theta_1=\frac\lambda d\)
This indicates that the path difference between wavelets from A and B is λ and therefore the wavelets from the corresponding points of the two halves AC and CB will have a path difference of \(\fracλ2\)
For the second minima: n = 2, \(\Rightarrow\theta_2=\frac{2\lambda}d\)
For the nth minima: n = n, \(\Rightarrow\theta_n=\frac{n\lambda}d\)
- For the nth secondary maxima (bright fringe), we have
Path difference = \(\left(2n+1\right)\frac\lambda2\)
\(\Rightarrow d\sin\left(\theta_n\right)=\left(2n+1\right)\frac\lambda2\)
\(\Rightarrow\sin\left(\theta_n\right)=\frac{\left(2n+1\right)\lambda}{2d}\)
Since θ is small, sin(θ) ≅ θ
∴ \(\theta_n=\frac{\left(2n+1\right)\lambda}{2d}\)
For the first order maxima, n = 1 \(\Rightarrow\theta_1=\frac{3\lambda}{2d}\)
For the second maxima, n = 2 \(\Rightarrow\theta_2=\frac{5\lambda}{2d}\)
For the nth maxima, n = n \(\Rightarrow\theta_n=\frac{(2n+1)\lambda}{2d}\)
- For nth secondary minima, we have,
Hence the condition for secondary maxima is θn = \(\theta_n=\frac{(2n+1)\lambda}{2d}\) and that of the secondary minima is θn = \(\theta_n=\frac{n\lambda}d\)