The Spot of Arago

Choose an opaque disc, a round hole or a zone plate, the wavelength and the distance to the screen. The page computes the light on the screen by Fresnel diffraction, ring by ring, draws the Fresnel zones on the obstacle, and leaves ray optics' answer as a dashed ghost.

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Lesson

The theory — The Spot of Arago

Seen from a point a distance z behind a plane, the plane divides into Fresnel zones, rings with edges at rn = √(nλz), the light from each arriving half a wavelength behind the last. A disc or hole of radius a covers N = a²/λz of them. In Fresnel's theory the centre of an opaque disc's shadow is exactly as bright as the open beam for every N: this is the spot of Arago, also called Poisson's spot. The centre of a hole's light is 4 sin²(πN/2) times the open beam, black whenever N is even. A zone plate blocks every other zone and focuses light at f = r1²/λ.

What each symbol means

a
the radius of the disc or hole
z
the distance from it to the screen
λ
the wavelength
N
a²/λz, the number of Fresnel zones the disc or hole covers
r_n
the outer edge of the nth zone, √(nλz); on a zone plate, r1√n
Breaks when
Fresnel's theory assumes the light reaches the screen at small angles. The exact scalar theory puts the centre of the shadow at z²/(z² + a²) of the open beam: 0.999996 for Arago's screen at half a metre, but 0.8 at z = 2a and a half at z = a. Fresnel stated the same proviso: the theorem holds “tant qu'il n'en résulte pas une trop grande obliquité”.

The centre of a shadow is bright 🖖

Press Arago's 2 mm screen. Ray optics draws a black circle, the dashed ghost. The computed shadow is dark, under a fifth of the open beam everywhere from the spot out to nine-tenths of the way to the edge, yet at the very centre it reads 1.000: as bright as if the disc were not there. Poisson found this in the integrals of Fresnel's memoir for the Académie's 1818 prize and offered it as an objection. Arago's report, read on 15 March 1819, says “l'observation a parfaitement confirmé le calcul”, and Fresnel credits Arago with seeing it behind a screen 2 millimetres across. The inset shows why it happens. Seen from the centre of the screen, the plane of the disc divides into Fresnel zones, rings whose edges are √(nλz) out, the light from each arriving half a wave behind the last. The disc blocks the first 3.160 of them, and everything outside it adds up to light exactly as bright as the open beam, whatever the disc's size or distance, as long as the light bends through small angles: the table below runs radii from 0.5 to 5 mm at 5 cm and at 1 m, and the numerical integral gives 1.0000 every time. Fresnel added that the result does not depend on the wavelength, so the spot must be white in white light. Press The same in white light: its colour differs from the open beam's by Δu′v′ = 0.0000. A hole is different, as Poisson also saw: press Poisson's second case: a hole, which is black at its centre in red light and blue in white.

The spot needs a point of light and a true circle 🖖

Press Lit by the Sun. Each point of a source casts its own pattern, moved sideways by z times its angle, and the page adds them all up. With a source the size of the Sun, 4.65 mrad in radius, the centre falls from 1.000 to 0.018. The spot is a small picture of the source, about 0.383 λz/a in radius from a single point, and a source whose angular radius approaches 0.383 λ/a smears it away: behind the 2 mm disc the spot keeps half its height only while the source is under 0.169 mrad in radius, a pinhole 0.34 mm across at a metre, 27 times smaller than the Sun. Maraldi, in 1723, let sunlight into a dark room through a small hole and hung a ball one ligne across, about 2.3 mm, 25 pieds from it, about 8 m, and saw “au milieu une aire circulaire fort claire” in its shadow, larger behind smaller balls, as 0.383 λz/a says it should be. He explained it as light flowing round the ball. The edge matters as much as the source. The light from the edge reaches the centre with a phase πr²/λz, so an edge that wanders scrambles it. How much depends on how the errors are spread; for errors scattered like a bell curve with r.m.s. σ the spot falls as exp(−(2πaσ/λz)²), for Arago's screen at half a metre to half at σ = 42 µm. Behind a disc 60 µm across, rough to about 300 nm, a beam of deuterium molecules gave a spot about a hundredth of the open beam.

Block every other zone and the shadow becomes a lens 🖖

Press Zone plate, red focus. On its own, one Fresnel zone sends the centre twice the open beam's amplitude, and neighbouring zones arrive in opposite phase, which is why the whole open plane gives only half the amplitude of the first zone alone. Block the even zones and the odd ones all arrive in step: ten open zones give (2 × 10)² = 400 times the open beam, and the integral gives 399.999 at 0.3949 m. The edges rn = r1√n are cut into the plate, so they match the Fresnel zones one for one only at the distance f = r1²/λ, which is inversely proportional to the wavelength: 0.3846 m for red 650 nm light and 0.5556 m for blue 450 nm, 1.44 times further. A glass lens does the opposite and brings blue to a focus nearer than red. The lower panel has spikes just as tall at f/3, f/5 and f/7, where each open ring holds three, five or seven Fresnel zones and the odd ones still outnumber the even. Set λ to 450 without moving the screen and the centre drops to 0.127. The zone plate is usually credited to J. L. Soret (1875); R. W. Wood's Physical Optics prints Rayleigh's notebook entry of 11 April 1871 describing one.

Problem solved in full

  1. A 2 cm disc, five metres from the wall 3 steps

    A disc 2 cm across stands 5 m from a wall, lit by a 633 nm laser. How many zones does it cover, how wide is the spot, and how smooth must its edge be, with errors scattered like a bell curve, before the spot loses half its brightness? The radius field stops at 5 mm, short of this disc. Press Arago's 2 mm screen first: it prints the spot's first minimum, 0.1205 mm, for a disc ten times smaller at a tenth of the distance.

    Open this problem in the tool

    1. The zones it covers, N = a²/λz:

    2. Deep in the shadow the spot's radius to its first minimum is 0.38274 λz/a, and λz/a is the same as for the 2 mm screen at half a metre:

    3. The edge error that halves the spot solves exp(−(2πaσ/λz)²) = ½:

    Answer

    31.6 zones, a spot 0.1211 mm in radius, and an edge true to 42 µm r.m.s. The deep-shadow formula gives the 2 mm screen the same 0.1211 mm, because it depends on a and z only through λz/a; the page's full integral puts that screen's first minimum at 0.1205 mm, since 3.16 zones is not yet deep shadow. Behind the 2 cm disc, with ten times the zones, the full integral and the formula agree to four places. The tolerance on the edge is set by λz/a too, so it is no tighter, although the shadow round the spot is now far darker.

Learning path

Wave optics

References (3)