Question

Explain Fraunhofer diffraction at a single slit with a neat ray diagram. Obtain an expression for the width of the central bright fringe. 

Answer 1

Fraunhofer diffraction due to single slit:

1. Consider a narrow slit AB of width ‘a’, kept perpendicular to the plane of the paper. The slit can be imagined to be divided into extremely thin slits or slit elements. It is illuminated by a parallel beam of monochromatic light of wavelength λ i.e., a plane wavefront is an incident on AB.
2. The diffracted light is focused by a converging lens L, on a screen XY.
3. The screen is kept in the focal plane of the lens and is perpendicular to the plane of the paper.
4. Let D be the distance between the slit and the screen.
5. According to Huygens’ principle, each and every point of the slit acts as a source of secondary wavelets, spreading in all directions.

Position of central maxima:

1. Let C be the centre of the slit AB. The secondary wavelets traveling parallel to CP_o come to a focus at P_o. The secondary wavelets from points equidistant from C in the upper and lower halves of the slit travel equal paths before reaching P_o. 
2. The optical path difference between all these wavelets is zero and hence they interfere in the same phase forming a bright image at P_o.
3. The intensity of light is maximum at the point P_o. It is called the central or the principal maxima of the diffraction pattern. 
4. For the line CP_o, angle θ = 0°.  

Position of secondary minima: 

1. Consider a point P on the screen at which waves travelling in a direction making an angle θ with CP are brought to focus at P by the lens. This point P will be of maximum or minimum intensity because the waves reaching P will cover the unequal distance.
2. Draw AN perpendicular to the direction of diffracted rays from point A. BN is the path difference between secondary waves coming from A and B.
3. From ΔABN, sin θ = ${\displaystyle \frac{\text{BN}}{\text{AB}}}$
   ∴ BN = AB sin θ = a sin θ
   Since θ is very small
   ∴ sin θ ≈ θ
   ∴ BN = aθ 
   In figure, suppose BN = λ and θ = θ₁ then sin θ₁ = ${\displaystyle \frac{\lambda }{\text{a}}}$ 
4. Such a point on the screen will be the position of the first secondary minimum. It is because, if the slit is assumed to be divided into two equal halves AC and BC, then the waves from corresponding points of two halves of the slit will have a path difference of λ/2.
   It gives rise to destructive interference at P which has minimum intensity.
5. If point P on the screen is such that BN = 2λ and angle θ = θ₂, then, sin θ₂ = ${\displaystyle \frac{2\lambda }{\text{a}}}$. Such a point on the screen will be the position of the second secondary minimum.
   In general, for n^th minimum, sinθ_n = ${\displaystyle \frac{\text{n}\lambda }{\text{a}}}$
   where, n = ±1, ±2, ±3,… 
6. If y_nd is the distance of n^th minimum from P_o, on the screen, then ${\displaystyle \left({\tan \theta }_{\text{nd}}\right)=\frac{{\text{y}}_{\text{nd}}}{\text{D}}}$.
7. If θ_nd is very small,
8. tan θ_nd ≈ sin θ_nd = ${\displaystyle \frac{\text{n}\lambda }{\text{a}}}$  
9. ∴ ${\displaystyle \frac{{\text{y}}_{\text{nd}}}{\text{D}}=\frac{\text{n}\lambda }{\text{a}}}$
10. ∴ ${\displaystyle {\text{y}}_{\text{nd}}=\frac{\text{n}\lambda \text{D}}{\text{a}}}$ = nW ....(1)
    where, W is fringe width Equation (1) gives distance of n^th secondary minima from central maxima. 
11. The central bright fringe is spread between the first dark fringes on either side. Hence, width of the central bright fringe is the distance between the centres of first dark fringe on either side.
    ∴ Width of the central bright fringe,
    W_c = 2y_1d = 2W = 2${\displaystyle \left(\frac{\lambda \text{D}}{\text{a}}\right)}$
    This is the required expression for the width of the central bright fringe.