Diffraction
The single slit
Look very closely at the shadow of a sharp edge, like a razor blade, in a narrow beam of light. The edge of the shadow is not sharp. Near it there are faint bright and dark bands, and a little light bends into the shadow. This is diffraction.
Diffraction happens with all waves. Sound bends easily round a door, because its wavelength is large. The wavelength of light is so small that we notice its bending only at very small openings or edges.
Send parallel light of wavelength λ through a single narrow slit of width a, and look at a distant screen. There is a broad, bright central maximum, with weaker bright and dark bands on both sides.
By Huygens’ principle, think of the slit as a row of many tiny sources. Straight ahead, at θ = 0, all of them are the same distance from the screen, so their waves arrive in phase. The centre is brightest.
At an angle θ, the waves from the two edges of the slit, M₁ and M₂, travel different distances. Their path difference is a sin θ, or about aθ for small angles.
Now let aθ = λ. Divide the slit into two halves. Every point in the upper half has a partner in the lower half, a/2 below it, whose path differs by λ/2. Each pair cancels, so the intensity is zero at θ = λ/a.
In the same way, the intensity is zero at θ = nλ/a, for n = ±1, ±2, ±3 and so on. Divide the slit into 2n equal parts; neighbouring parts cancel in pairs.
At θ = 3λ/2a, divide the slit into three equal parts. Two of them cancel, and only one third of the slit is left to contribute. This gives a secondary maximum, much weaker than the centre. The maxima near θ = (n + ½)λ/a get weaker and weaker.
Compare with Young’s experiment. There, ideal fringes are equally bright and equally spaced. In single-slit diffraction, the central maximum is twice as wide as the others and much brighter, and the side maxima fade quickly.
The central maximum, and when ray optics works
The central maximum stretches from the first minimum on one side, θ = −λ/a, to the first minimum on the other side, θ = +λ/a. So its angular width is 2λ/a.
On a screen a distance D away, the first minima are at x = ±λD/a. So the central maximum is 2λD/a wide. A narrower slit gives a wider central maximum.
Example: light of wavelength 600 nm passes through a slit 0.2 mm wide, and the screen is 2 m away. The first minimum is at λD/a = (6 × 10⁻⁷ × 2)/(2 × 10⁻⁴) = 6 × 10⁻³ m, or 6 mm. The central maximum is 12 mm wide.
Real double slits have a width too. So Young’s fringes are really a double-slit interference pattern, whose brightness follows a single-slit diffraction envelope.
Example: slits 1 mm apart. How wide should each slit be so that about 10 bright fringes of the double slit fit inside the central maximum? We need 2λ/a = 10λ/d, so a = d/5 = 0.2 mm.
Diffraction also tells us when ray optics works. A beam through an aperture of size a spreads out at an angle of about λ/a. After a distance z, it has widened by about zλ/a.
This spreading becomes as large as the aperture itself when zλ/a = a, that is, at z = a²/λ. This distance is called the Fresnel distance, z_F = a²/λ.
For distances much smaller than z_F, the spreading is small compared with the size of the beam, and light travels in straight lines, as rays. Ray optics is the limit of wave optics as the wavelength tends to zero.
Example: an aperture 3 mm wide, with light of wavelength 500 nm. z_F = (3 × 10⁻³)²/(5 × 10⁻⁷) = (9 × 10⁻⁶)/(5 × 10⁻⁷) = 18 m. Even for a small aperture, ray optics works well over many metres.
Try it: single-slit diffraction
- Drag P to the first dark band. The readout shows a x/D = λ: the two halves of the slit cancel.
- Halve the slit width a: the central maximum doubles in width. Check that its width is 2λD/a.
- Change λ from violet to red: red light diffracts more. Tick “White light” to see coloured edges.
- Read the Fresnel distance a²/λ for a 0.2 mm slit, then for 1 mm. Wider apertures behave like rays for much longer.
Width of the central maximum
Light of wavelength 500 nm falls on a slit 0.1 mm wide. The screen is 1 m away. How wide is the central maximum on the screen?
- 5 mm
- 10 mm
- 1 mm
- 0.5 mm
Answer
B. The first minima are at x = ±λD/a = ±(5 × 10⁻⁷ × 1)/(1 × 10⁻⁴) = ±5 × 10⁻³ m. The central maximum goes from one to the other, so its width is 2λD/a = 10 mm. (5 mm is only half of it, from the centre to one minimum.)
Fresnel distance
Light of wavelength 400 nm passes through an aperture 2 mm wide. Up to roughly what distance can the beam be treated with ray optics?
- 5 mm
- 1 m
- 10 m
- 5000 m
Answer
C. The Fresnel distance is z_F = a²/λ = (2 × 10⁻³)²/(4 × 10⁻⁷) = (4 × 10⁻⁶)/(4 × 10⁻⁷) = 10 m. Well within this distance, diffraction spreading is small compared with the beam. (5000 m comes from a/λ, without squaring a.)
Summary
- Path difference between the edges:
a sin θ ≈ aθ - Minima:
θ = nλ/a (n = ±1, ±2, …) - Central maximum on a screen at D:
width = 2λD/a - Fresnel distance:
z_F = a²/λ
- Diffraction is the bending of waves around edges and through openings; it is noticeable when the opening is not much larger than λ.
- A single slit gives a bright central maximum, with dark bands at θ = nλ/a and much weaker secondary maxima near θ = (n + ½)λ/a.
- A narrower slit or a longer wavelength gives a wider central maximum.
- Double-slit fringes are equally bright only for very narrow slits; real fringes lie inside a single-slit envelope.
- For distances much less than z_F = a²/λ, diffraction spreading is negligible and ray optics is valid.