Question

Derive the conditions for bright and dark fringes produced due to diffraction by a single slit.

Answer 1

When a single slit of finite width is illuminated by a parallel beam of monochromatic light of wavelength λ, we see a broad pattern of alternate dark and bright fringes on a screen some distance from the slit. The pattern is made up of a central bright fringe, followed by dark and bright fringes of diminishing intensity on both sides. This is called the diffraction pattern of a single slit. Consider a single slit that is illuminated by a monochromatic light beam that is parallel to the slit's plane. The diffraction pattern is obtained on a screen at a distance D ( » a) from the slit and at the focal plane of the convex lens,

Fraunhofer diffraction due to a single slit

We can imagine the single slit as being made up of a large number of Huygens' sources that are evenly distributed across its width. Then the maxima and minima of the pattern arise from the interference of the various Huygens' wavelets. Consider the single slit as being made up of two adjacent slits, each with a/2 width. All Huygens sources at the slit will be in phase because the incident plane wavefronts are parallel to the slit plane.

They will therefore also in phase at the point P₀ on the screen, where P₀ is equidistant from all the Huygens sources. At P₀ then, we get the central maximum.

For the first minimum of intensity on the screen, the path difference between the waves from the Huygens sources A and O (or 0 and B) is λ/2, which is the condition for destructive interference. Suppose, the nodal line OP for the first minimum subtends an angle θ at the slit; θ is very small. With P as the centre and PA as radius, strike an arc intersecting PB at C. Since, D » a, the arc AC can be considered a straight line at right angles to PB. 
Then, Δ ABC is a right-angled triangle similar to Δ OP₀P.

This means that, ∠ BAC = θ

∴ BC = a sin θ

∴ Difference in path length,

BC = PB - PA = (PB - PO) + (PO - PA)

=`lambda/2 + lambda/2 = lambda`

∴ a sin θ =  λ

∴ sin θ ≅ θ = `lambda/"a"`     .....(1)

(∵ θ is very small and in radian)

The other nodal lines of intensity minima can be understood in a similar way. In general, then, for the mth minimum (m = ± 1, ± 2, ± 3,....)

`theta_"m" = ("m"lambda)/"a"` (mth minimum) ...(2)

as θ_m is very small and in radian.

Between the successive minima, the intensity rises to secondary maxima when the path difference is an odd-integral multiple of `lambda/2`:

a sin θ_m = (2m + 1)`lambda/2 = ("m" + 1/2) lamda`

i.e., at angles given by,

θ_m ≅ sin θ_m = `("m" + 1/2)lambda/"a"`

(mth secondary maximum)        ......(3)