Question

(a) State Huygen’s principle. Using this principle explain how a diffraction pattern is obtained on a screen due to a narrow slit on which a narrow beam coming from a monochromatic source of light is incident normally. 

(b) Show that the angular width of the first diffraction fringe is half of that of the central fringe. 

(c) If a monochromatic source of light is replaced by white light, what change would you observe in the diffraction pattern?

Answer 1

(a) Huygen’s principle is useful for determining the position of a given wavefront at any time in the future if we know its present position. The principle may be stated in three parts as follows: 

(i) Every point on a given wavefront may be regarded as a source of new disturbance. 

(ii) The new disturbances from each point spread out in all directions with the velocity of light and are called the secondary wavelets. 

(iii) The surface of tangency to the secondary wavelets in forward direction at any instant gives the new position of the wavefront at that time. 

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,

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 

a sin θ= nλ...(iii) 

Secondary Maxima: Let us now consider angle θsuch that

(b) Angular width of central maximum (β_θ)_c = 2λ/a

Angular width of first maximum, ( β_I) = 2λ/a - λ/a = λ/a

β_I/(β_θ)_c = 1/2

Hence, the fringe width of the first diffraction fringe is half that of the central fringe. 

(c) If monochromatic light is replaced by white light, each diffraction band gets splited into the number of coloured bands, the angular width of violet is least and that of red is maximum.