Question

(a) Discuss the diffraction produced by a narrow slit which is illuminated by monochromatic light.

(b) Show that the central maximum in the single slit diffraction is twice as wide as the second maxima and the pattern becomes narrow as the width of slit is increased.

Or

Using Huygens’ principle, draw a diagram to show propagation of wavefront originating from a monochromatic point source.

Or

Describe the diffraction of light due to a single slit. Explain fromation of a pattern of fringes obtained on the screen and plot showing the variation of intensity with angle 6 in single slit diffraction.

Answer 1

Huygens’ principle and the diagram for propagation of w avefront.

Assumptions on which Huygens’ principle is based are:

• The medium is homogeneous.
• The medium is isotropic.

Huygens’ principle

(a) Each point of a wavelength becomes a source of new disturbance called the secondary wavelets which travel in all directions with the same speed provided the medium remains the same.

(b) The secondary wavefront is the tangent plane joining all the wavefronts.

Fig. (a) represents the trace of secondary wavefront at a certain instant and Fig. (b) represents the trace when the light has travelled a large distance.

Rays of light are perpendicular to the wavefront.

(a) Diffraction of light by narrow slit

Consider a narrow slit AB perpendicular to plane of paper on which a monochromatic light be incident normally. Due to diffraction pattern consists of a central band much wider than the slit width just opposite to the slit and bordered on either side by dark and bright fringes of decreasing intensity. It is found that

• the intensity of the secondary maxima goes on decreasing.
• the width of central maximum is double that of a secondary maximum.

Consider a plane wavefront WW incident on the slit AB (Fig. Imagine the slit to be divided into a large number of very narrow strips of equal width parallel to the slit. When the wavefront reaches the slit, each narrow strip parallel to the slit can be considered to be a source of Huygens’ secondary wavelets.

Central Maxima: Let us first consider the effect of all the wavelets at the point O just opposite to the middle of the slit. Since distance between the slit and the screen is very large as compared wdth the width of the slit, the wavelets cover the same distance and reach O in the same phase. Thus the wavelets reinforce each other’s effect to give maximum intensity at O.

Position of secondary minima

Let us now consider the intensity at a point P above O produced by the rays travelling at an ∠θ with the original direction. Since the width of the slit is very small, all the rays reaching P proceed almost parallel to each other. The wavelets from all the points in the slit start in the same phase but as they travel different distances, they reach P in different phases. Draw AL normal to BP.

The path difference between the extreme rays is BP - AP = BL.

Let C be the mid-point of slit or path difference between AP and CP = \(\frac{\lambda}{2}\)

∴ Wavelets starting from A and C shall reach the point P in opposite phases and cancel each other’s effect.

Similarly, the wavelets from a strip below A are cancelled by the wavelets from a corresponding strip below C.

In ∆ ABL, 

\(\frac{BL}{AB} \) = sin θ

∴ BL = AB sin θ

or BL = d sin θ

For first minima, BL = λ;

∴ d sin θ = λ

or sin θ = 

Since θ is small ∴ sin θ = θ = \(\frac{x}{D}\)

\(\frac{x}{D} = \theta = \frac{\lambda}{d}\) or x = \(\frac{D}{d} \lambda\)

At a point slightly above P i.e. at P₁, the path difference between the two rays in BL₁.

Position of secondary maxima

If BL₁ = \(\frac{3}{2}\lambda\) then slit AB can be divided into three equal parts Fig. (b). The waves from corresponding points AC₁ and C₁C₂ cancel each other. The waves from C₂B produce first secondary maximum at an angle θ₁ given by

d sin θ₁ = \(\frac{3}{2}\lambda\)

sin θ₁ = \(\frac{3}{2} \frac{\lambda}{d}\)

If θ₁ is small then sin θ₁ = θ₁

It another point P₂, Fig. (c), the path difference between two rays is BL₂ = 2λ, then, the slit AB can be supposed to be divided into four equal parts,. The waves from corresponding points of the parts AC₁ and C₁C₃ cancel each other. Also the waves from the corresponding points C₂C₃ and C₃B cancel each other producing second minima

The intensity of central fringe is maximum whereas that of other fringes falls off rapidly in either direction from the centre of the fringe pattern.

(b) Width of central maxima is the distance between first secondary minimum on either side of central point in front of slit.

∴ d sin θ = 1λ for first secondary minima.

or sin θ = \(\frac{\lambda}{d}\)

If f is the focal length of the focussing lens (on screen side), held close to the slit, then D, the distance of slit from the screen is equal to f i.e., D = f

∴ sin θ ≈ θ = \(\frac{y}{f} = \frac{y}{D}\)

or \(\frac{y}{D} = \frac{\lambda}{d}\) or y = \(\frac{D\lambda}{d}\)

Since θ is half the angular width of central maxima of single slit,

∴ Width of central maxima

 = 2y = \(\frac{2D\lambda}{d} = \frac{2f}{d}\) ..........(i)

Width of secondary maxima is the distance between nth and (n + 1)^th maxima.

We know for a nth minima a sin θ_n = nλ

Thus we find that equation

• is twice the equation
• showing thereby that the central maxima is twice any secondary maxima.

If the width of slit is increased then as y decreases which shows that pattern will become narrower as the width of the slit is increased.