On a dark night, look at a distant street light through a thin, finely woven dupatta held close to your eye. The single point of light becomes a small cross. Along each arm of the cross sit fainter spots of light, evenly spaced. Stretch the cloth, and the pattern changes. Take the cloth away, and there is only the point again.
Sound bends round a door
Stand beside an open door and someone in the next room can still call you. The sound reaches you although there is no straight line from the speaker to your ear. A note of \(340\,\text{Hz}\) in air has a wavelength of \(340/340 = 1\,\text{m}\), about the width of the door. Waves of that size spread out widely as they pass the opening.
Light does not seem to do this. The edge of a shadow in sunlight looks sharp, and you cannot see round a door. The chapter on ray optics built everything on this: light travels in straight lines. Yet light is a wave, as the chapter on Young's double slit showed. A wave should bend round an edge too.
The answer is size. Visible light has a wavelength of about \(500\,\text{nm}\), two million times shorter than the sound. An opening of a metre is enormous compared with that. Light does bend round every edge, but through an angle too small to notice unless the opening is very fine.
Diffraction is the bending of a wave round the edges of an obstacle or an opening, into the region of the geometric shadow. It is noticeable only when the size of the opening or obstacle is comparable with the wavelength.
Wavelets from every point of the opening
Huygens' principle, from the chapter on Young's double slit, explains the bending. Every point of a wavefront is a source of secondary wavelets. When a straight crest reaches an opening, only the points inside the opening send wavelets onward. The new crest beyond the opening is the envelope of those wavelets.
Across the middle of a wide opening, the wavelets build a straight crest again, and the wave goes on as a beam. Only near the two edges does the envelope curve round into the shadow. Through a narrow opening, the few wavelets make almost a half circle. The wave then spreads to every side.
In Fig 1.1, a gap five wavelengths wide bends the crests only at their ends, up to \(11.5°\). A gap two wavelengths wide spreads them to \(30°\). A gap of one wavelength turns each crest into a half circle. The narrower the gap compared with \(\lambda\), the wider the wave spreads.
The spread is about \(\lambda/a\) in radians, where \(a\) is the width of the opening. For light of \(500\,\text{nm}\) through a window \(1\,\text{m}\) wide, \(\lambda/a = 5 \times 10^{-7}\,\text{rad}\). That is far too small to see. Through a slit \(0.1\,\text{mm}\) wide, \(\lambda/a = 5 \times 10^{-3}\,\text{rad}\). On a wall \(1\,\text{m}\) away, the light then spreads over about \(5\,\text{mm}\), easily seen.
A shadow is sharp for the same reason that a large ship ignores small ripples. Waves much shorter than an object pass it almost as rays. Ray optics is the limit of wave optics when the wavelength is tiny beside every opening.
Light passes through an open window \(1\,\text{m}\) wide, and also through a slit \(0.1\,\text{mm}\) wide. Why is diffraction seen clearly only with the slit?
Francesco Maria Grimaldi, a Jesuit priest, taught mathematics at Bologna in Italy. He let sunlight into a dark room through a tiny hole and held a thin rod in the beam. The shadow of the rod was wider than straight rays allowed. Its edges were fringed with coloured bands. Grimaldi called the effect diffractio, a breaking apart of light. His book was published in 1665, two years after his death.
A single slit, lit by parallel light
Shine parallel light of one wavelength on a long, narrow slit of width \(a\). Look at the light that leaves it at an angle \(\theta\) to the straight-through direction. Either catch it on a screen far away, or bring it to a focus with a convex lens. In both cases, the rays that reach one point P of the screen leave the slit parallel. This arrangement is Fraunhofer diffraction, and it is the one the syllabus uses.
Straight ahead, at \(\theta = 0\), every wavelet from the slit has travelled the same distance to the screen. All of them arrive in phase and add up. The centre of the screen is therefore bright, the brightest point of the pattern.
Now tilt the direction to an angle \(\theta\). Call the edges of the slit A and B. Draw AN at right angles to the rays, as in Fig 1.2. From the line AN onward, every ray travels the same distance to P. So the ray from B travels an extra distance BN, and \[ BN = a\sin\theta. \] The rays from points between A and B lag by amounts between zero and \(a\sin\theta\).
Pairing the two halves
Suppose \(a\sin\theta = \lambda\), the first chip of Fig 1.2. Split the slit into an upper half and a lower half. Take a point in the upper half, and its partner \(a/2\) below it in the lower half. Their rays differ in path by \(\tfrac{a}{2}\sin\theta = \lambda/2\). They arrive at P crest on trough and cancel.
Every point of the upper half has such a partner, and every pair cancels. So the whole slit sends no light to P, and P is dark. This is the first minimum of the pattern.
The same pairing works whenever the slit splits into an even number of zones. At \(a\sin\theta = 2\lambda\), the third chip, cut the slit into four quarters. Neighbouring quarters cancel pair by pair, and P is dark again. In general, P is dark whenever \(a\sin\theta\) is a whole number of wavelengths.
\[ BN = a\sin\theta \]
Why. AN is at right angles to the rays, so beyond AN every ray travels the same distance. B's ray travels the extra BN, a side of the right triangle ABN with hypotenuse \(a\) and angle \(\theta\).
\[ \text{points } \tfrac{a}{2} \text{ apart differ in path by } \tfrac{a}{2}\sin\theta \]
Why. the extra path grows in proportion to the distance down the slit.
\[ \tfrac{a}{2}\sin\theta = \tfrac{\lambda}{2} \;\Rightarrow\; a\sin\theta = \lambda \]
Why. a path difference of half a wavelength makes each pair arrive in opposite phase, so each pair cancels.
\[ \tfrac{a}{2n}\sin\theta = \tfrac{\lambda}{2} \;\Rightarrow\; a\sin\theta = n\lambda \]
Read it. with \(2n\) equal zones, neighbouring zones cancel pair by pair in the same way.
For parallel light of wavelength \(\lambda\) on a slit of width \(a\), the light is dark in the directions \(\theta\) that satisfy \(a\sin\theta = n\lambda\), with \(n = \pm1, \pm2, \pm3, \dots\)
The formula looks like the one for the bright fringes of Young's double slit, \(d\sin\theta = n\lambda\). Here it gives the dark bands. Also, \(n = 0\) is not a minimum: at \(\theta = 0\) all the wavelets are in phase, and that is the brightest point.
Parallel light leaves a slit of width \(a\) at an angle \(\theta\). What is the path difference between the rays from the two edges of the slit?
Light of wavelength \(\lambda\) falls on a slit of width \(a = 4\lambda\). In which of these directions is the screen dark?
How wide the central band is
On a screen at a distance \(D\), the first minimum at the angle \(\theta_1\) lies a distance \(y_1 = D\tan\theta_1\) from the centre. For light the angle is small, so \(\sin\theta \approx \tan\theta \approx \theta\) in radians. Then \(\theta_1 = \lambda/a\), and \[ y_1 = \frac{\lambda D}{a}. \]
The central maximum is the bright band between the first minimum on one side and the first minimum on the other. So its width is twice \(y_1\): \[ w = \frac{2\lambda D}{a}. \] Its angular width is \(2\lambda/a\). If a convex lens of focal length \(f\) focuses the light instead, put \(f\) in place of \(D\).
A slit \(0.2\,\text{mm}\) wide is lit by parallel light of \(500\,\text{nm}\). The screen is \(2\,\text{m}\) away. Find the angle of the first minimum, its distance from the centre, and the width of the central maximum.
\(a = 0.2 \times 10^{-3}\,\text{m}\), \(\lambda = 500 \times 10^{-9}\,\text{m}\), \(D = 2\,\text{m}\).
First minimum: \(\sin\theta_1 = \lambda/a\), and \(y_1 = D\theta_1\) for a small angle.
\(\theta_1 = (500 \times 10^{-9})/(0.2 \times 10^{-3}) = 2.5 \times 10^{-3}\,\text{rad}\), and \(y_1 = 2 \times 2.5 \times 10^{-3} = 5 \times 10^{-3}\,\text{m}\).
The first minimum is \(5\,\text{mm}\) from the centre, and the central maximum is \(2y_1 = 10\,\text{mm}\) wide.
The formula \(w = 2\lambda D/a\) says three things at once. Double the width of the slit, and the central band halves: a wider slit gives a narrower pattern. Double the wavelength, and the band doubles, so red light spreads more than blue. Move the screen farther away, and the band grows in proportion.
Light of \(500\,\text{nm}\) falls on a single slit. On a screen \(1\,\text{m}\) away, the first minimum is \(2\,\text{mm}\) from the centre. Find the width of the slit, in mm.
A slit \(0.1\,\text{mm}\) wide is lit by light of \(600\,\text{nm}\). How wide is the central maximum on a screen \(1.5\,\text{m}\) away, in mm?
Which of these changes make the central maximum of a single slit wider?
Now return to the dupatta. Its threads leave tiny openings, roughly \(0.1\,\text{mm}\) across. Each opening spreads the light of the street lamp through an angle of about \(\lambda/a = 5 \times 10^{-3}\,\text{rad}\), close to \(0.3°\). That is half the width of the full Moon, just large enough to see. The threads run two ways, warp and weft, so the light spreads along two lines at right angles: a cross.
The separate spots along each arm come from the many openings together. Openings repeated at a regular spacing \(d\) send light that reinforces only in the directions where \(d\sin\theta = n\lambda\). Such a regular array is a diffraction grating. Stretching the cloth changes \(d\), and the spots move.
- Thinking that light does not diffract through a window because windows are made of glass.Light diffracts at every edge. Through a 1 m window the angle \(\lambda/a\) is only \(5 \times 10^{-7}\,\text{rad}\), far too small to see.
- Using \(a\sin\theta = n\lambda\) for the bright bands of a single slit.For a single slit this condition gives the dark bands. The bright bands lie roughly midway between them.
- Taking \(\lambda D/a\) as the width of the central maximum.\(\lambda D/a\) is the distance from the centre to the first minimum. The central band spans both sides, \(2\lambda D/a\).