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Ibn al-Haytham

Abū ʿAlī al-Ḥasan ibn al-Ḥasan ibn al-Haytham (أبو علي الحسن بن الحسن بن الهيثم) · Latin Alhacen, Alhazen

A course in two strands: the life of a scholar from Basra who served, and then fled, a Fatimid caliph in Cairo, and the optics, experimental method, astronomy and mathematics with which he rebuilt the science of light and questioned Ptolemy.

Born c. 965, Basra Died c. 1040, Cairo Book Kitāb al-Manāẓir (Book of Optics), seven books Level advanced undergraduate
Alhazen’s problem in a circular mirror. From A, four rays reflect to B. All four reflection points lie on one rectangular hyperbola through the centre O, a modern form of the conic solutions of Christiaan Huygens.
Lesson 00

How to use this course

Ibn al-Haytham wrote the most important work on light between antiquity and the seventeenth century. His Book of Optics argued that we see because light enters the eye, gave the law of reflection a complete form, posed and solved a hard geometric problem about mirrors, and treated refraction by measurement. It did all this through controlled tests with purpose-built apparatus, which he called iʿtibār. In Latin translation the book shaped optics in Europe from Roger Bacon to Johannes Kepler. He also wrote a sharp critique of Ptolemy’s astronomy and some fine pieces of mathematics.

His life is known from a short autobiography, three lists of his works, and stories told by biographers who wrote a century or two after his death. The stories disagree. Part I sets out what each source says. Part II follows the optics book by book, then turns to astronomy and mathematics. The exercises use modern notation, but most of them reconstruct arguments he made.

Prerequisites for Part II
  • Euclidean geometry: similar triangles, inscribed angles, tangents to circles and conics.
  • Trigonometry and complex numbers in polar form.
  • Some calculus, to compare his volume of the paraboloid with an integral.
  • For Lesson 09, the equant model from Lesson 08 of the Ptolemy course.

The course on Claudius Ptolemy gives the astronomy and optics he criticized. The course on Nasir al-Din al-Tusi follows the next stage of the critique of Ptolemy, at the Maragha observatory, and the course on Nicolaus Copernicus shows where it led.

Reference

Timeline

Dates in Ibn al-Haytham’s life come from the Islamic (hijri) calendar. Where a hijri year spans two Western years, the course gives the approximate one. Birth and death are known only roughly.

c. 965Born in Basra, in southern Iraq, then under the Buyid emirs.
969The Fatimids conquer Egypt and found Cairo.
c. 984Ibn Sahl writes On Burning Mirrors and Lenses, with the law of refraction in sine form (Rashed’s dating).
996Al-Ḥākim becomes Fatimid caliph at the age of eleven.
996–1021At some point in al-Ḥākim’s reign, Ibn al-Haytham goes to Egypt. The expedition to regulate the Nile fails.
c. 1011–21Traditional date range for the Kitāb al-Manāẓir.
1021Al-Ḥākim disappears and is presumed dead. According to Ibn al-Qifṭī, Ibn al-Haytham ends his pretended madness.
1024Dated copy, in his hand, of Apollonius’s Conics.
1025–28Al-Shukūk ʿalā Baṭlamyūs (Doubts concerning Ptolemy).
1027Autobiography and first list of works (to 10 February 1027).
1028Second list of works (to 25 July 1028); treatise on the halo and the rainbow.
1038Third list, of 92 titles (to 2 October 1038). The Model of the Motions of Each of the Seven Planets.
c. 1040Dies in Cairo. A geometry volume in his hand is dated 432 AH (1040–41).
c. 1200Latin translation of the Optics, De aspectibus or Perspectiva, by an unknown translator.
13th c.Roger Bacon, John Pecham and Witelo build Latin optics on it.
early 14th c.Kamāl al-Dīn al-Fārisī’s commentary Tanqīḥ al-Manāẓir revives the work in the Islamic world.
1572Friedrich Risner prints the Latin text at Basel in Opticae thesaurus, with Witelo’s Perspectiva.
1604Kepler’s Ad Vitellionem paralipomena solves the problem of the retinal image.
1673Huygens and René-François de Sluse publish conic solutions of Alhazen’s problem.
1965–98Jack Elkin, then Peter Neumann, prove that the problem cannot in general be solved with straightedge and compass.
1989A. I. Sabra’s English translation of Optics I–III.
2015UNESCO’s International Year of Light marks the thousand years since the Optics.
Part I · Formation
Lesson 01

Basra and the search for truth

Al-Ḥasan ibn al-Ḥasan ibn al-Haytham was born around 965 in Basra, the port city at the head of the Persian Gulf. The year comes from his own remarks and from the dates in his lists of works, which imply that he was in his sixties in the 1020s. Arabic writers called him al-Baṣrī, “of Basra”, and later al-Miṣrī, “the Egyptian”. Medieval Latin readers knew him as Alhacen; the form Alhazen became common after the printed edition of 1572. Whether his family was Arab or Persian is not recorded.

Basra was then part of the Buyid emirate, under the nominal authority of the Abbasid caliph in Baghdad. It was a commercial city with a long tradition of scholarship in grammar, theology and the sciences translated from Greek. By the tenth century Arabic readers had Euclid, Apollonius, Archimedes, Ptolemy, Aristotle and Galen, and a century of Arabic work that extended them.

The autobiography

In 1027 he wrote an account of his intellectual life, modelled on Galen’s book about his own writings. It says little about events. He describes how, as a young man, he studied the claims of the religious sects and found that each held its beliefs on authority. He resolved to accept only what could be shown by demonstration about things perceived by the senses, and he found that method in the writings of Aristotle and in mathematics. The same commitment opens the Doubts concerning Ptolemy, where he says that the seeker after truth must treat what he reads as an opponent, testing it from every side, and must suspect his own judgement as well.

A career in administration

Ibn Abī Uṣaybiʿa, writing in the thirteenth century, says that Ibn al-Haytham held an administrative post in Basra, sometimes described as a ministry, and that he wished to leave it for study. Al-Bayhaqī and Ibn al-Qifṭī add that he became known for his skill in applied mathematics and engineering. These details come from later writers and cannot be checked against his own words.

Lesson 02

Cairo, the Nile and al-Ḥākim

The Fatimid dynasty, a rival caliphate, took Egypt in 969 and built Cairo beside the older city of Fustat. In 996 the eleven-year-old al-Ḥākim became caliph. He founded a library and academy, the Dār al-ʿIlm, in 1005 and supported the astronomer Ibn Yūnus, whose great tables, the Ḥākimī Zīj, bear the caliph’s name. He was also notorious for unpredictable decrees and executions.

The Nile project

Ibn al-Qifṭī (d. 1248), the fullest source, says that Ibn al-Haytham, while still in Iraq, claimed he could build works to regulate the Nile so that Egypt would have water in every season. Al-Ḥākim invited him to Egypt and sent him south with a team of engineers. When he reached the cataracts near Aswan he saw that the ancient Egyptians, whose monuments he passed, would have done it if it were possible. He concluded that the scheme could not work and admitted the failure to the caliph. Al-Ḥākim gave him a government post instead.

The pretended madness

Ibn al-Haytham feared the caliph. In Ibn al-Qifṭī’s account he pretended to be insane, his property was taken into custody, and he was confined to his house until al-Ḥākim’s disappearance in 1021. He then showed that he was sane, recovered his property, and lived near the Azhar mosque, earning his living by copying scientific books. Ibn al-Qifṭī’s informant, Yūsuf al-Fāsī, says he copied Euclid’s Elements, the “middle books” of mathematical astronomy and Ptolemy’s Almagest, and earned about 150 dinars a year.

The story is often told as the origin of the Optics: a scholar under house arrest, studying the light that entered his darkened room. No source says so. The date range 1011–1021 sometimes given for the book rests on that reading of the story, and it should be held loosely.

Lesson 03

The sources and their stories

Three kinds of evidence survive, and they are of very different quality.

  • His own writings. The autobiography of 1027 and three lists of works, ending on 10 February 1027, 25 July 1028 and 2 October 1038. The last list has 92 titles. Ibn al-Qifṭī gives 69 titles, and all but two appear in the 1038 list. At least 55 works survive. Two manuscripts in his own hand are dated, a copy of Apollonius’s Conics (1024) and a geometrical volume (1040–41).
  • Ibn al-Qifṭī, in his History of the Learned: the Nile expedition, the feigned madness in Cairo, the copying of books, and death around 430 AH (1038–39). The volume of 1040–41 shows that this date is too early.
  • Other biographers. Al-Bayhaqī (d. 1169–70) tells of a failed meeting with al-Ḥākim and a flight to Syria; Sabra judged this unsupported. Ibn Abī Uṣaybiʿa (d. 1270) places the feigned madness in Basra, as a way to escape his post there. A title in the lists implies a stay in Baghdad around 1027.

One man or two?

Ibn Abī Uṣaybiʿa gives the autobiography and the first two lists under the name Muḥammad ibn al-Ḥasan ibn al-Haytham, and the third under al-Ḥasan ibn al-Ḥasan. Roshdi Rashed argued that these were two people: a philosopher and physician named Muḥammad, and the mathematician al-Ḥasan who wrote the Optics. A. I. Sabra replied in two papers titled “One Ibn al-Haytham or two?” (1998 and 2003) that the evidence points to a single author whose name was garbled in transmission. Sabra’s view is widely accepted, and this course treats all the works as those of one person.

Reading the legends

The madness story has the shape of a moral tale: a learned man outwits a tyrant. It may still be true in outline. What the historian can say is limited: he was in Egypt in the reign of al-Ḥākim; the Nile scheme is reported by more than one source; he was in Cairo in later life; and he made his living, at least in part, by his pen.

Exercise 3.1A hijri year has 354 or 355 days. The year 430 AH began on 3 October 1038. Explain why a death “around 430 AH” and a manuscript dated 432 AH cannot both describe the same man, and which piece of evidence a historian should prefer.

Show solution
432 AH began about two lunar years after 430 AH, in September 1040, and ended in 1041. A man who died in 1038–39 could not date a manuscript in 1040–41. The dated autograph is primary evidence, a physical object; Ibn al-Qifṭī wrote about two centuries later from informants. The autograph wins, and the death is put at 1040 or soon after, with “c.” to mark the doubt.
Part II · The optics and mathematics
Lesson 04

Light, rays and the method of test

For the general reader. Greek optics was mostly a geometry of sight lines, drawn from the eye to the object. Ibn al-Haytham asked what light is and how it travels, and he answered by building apparatus: tubes, screens with holes, darkened rooms. He wanted every claim to be checked by an arrangement that anyone could repeat.

Goals
  • Know the plan of the seven books of the Optics.
  • State the principles of light in Book I and the tests that support them.
  • Analyse image formation through an aperture, the camera obscura.

The plan of the book

  • Book I. Light and colour; the anatomy of the eye; the mode of vision.
  • Book II. Perception: how the mind recognizes shape, size, distance and the other visible properties.
  • Book III. Errors of direct vision, including binocular vision.
  • Book IV. Reflection: its laws and the apparatus for testing them.
  • Book V. Images in plane, spherical, cylindrical and conical mirrors, including Alhazen’s problem.
  • Book VI. Errors of vision due to reflection.
  • Book VII. Refraction, with a refraction apparatus, and errors due to refraction.

The Arabic word iʿtibār, translated into Latin as experimentum, means testing or verification. The Optics uses it for a deliberate check, with an instrument, of a claim stated beforehand. The work cites no authority by name.

Principles of Book I (paraphrased)
  1. Light is a real property of bodies. Self-luminous bodies such as the Sun and fire have “essential” light; illuminated bodies have “accidental” light, which they in turn send out.
  2. From every point of a luminous or illuminated body, light goes out in straight lines in every direction where there is a transparent medium.
  3. Lights from different sources cross in the air without mixing or affecting one another.
  4. Colour travels with light and is seen only with it.

The tests

For straight-line propagation he looked through straight tubes, and placed straight rulers along beams entering a darkened room through a hole. For the third principle he described a test with several candles in a row outside a dark chamber. Each candle sends a separate spot of light through a single aperture onto the opposite wall, on the straight line through the hole. Cover one candle, and only its spot disappears. The lights have crossed at the aperture and passed through each other unchanged.

The dark chamber

He called it al-bayt al-muẓlim, the dark room. In a short treatise On the Shape of the Eclipse he discussed why, during a partial solar eclipse, sunlight through a small hole forms a crescent, and under what conditions the image through an aperture is distinct. His answer is incomplete, but it shows that he understood the principle, which can be put as follows.

Proposition (image through an aperture)

A luminous disc of angular diameter \(\alpha\) shines through an aperture of diameter \(w\) onto a screen at distance \(D\) behind it. Each point of the source projects a copy of the aperture, displaced along the straight line through the aperture. The illuminated patch is the union of these copies: an inverted image of the source of diameter about \(\alpha D\), blurred by the aperture width \(w\). When \(\alpha D\gg w\) the patch has the shape of the source; when \(\alpha D\ll w\) it has the shape of the aperture.

Exercise 4.1The Sun has angular diameter \(0.53^\circ\). A hole of 5 mm in a shutter casts light on a wall 4 m away. Find the diameter of the Sun’s image and decide whether a crescent will be visible during a partial eclipse. What if the wall is only 30 cm away?

Show solution
\(\alpha=0.53^\circ=0.00925\) rad, so \(\alpha D=37\) mm at 4 m, much larger than \(w=5\) mm: the image is a sharp enough disc, and a crescent shows. At 30 cm, \(\alpha D=2.8\) mm, smaller than the hole: the patch is roughly the shape of the hole and the crescent is lost. This is why sharpness requires the hole to be small relative to the distance.

Exercise 4.2In the candle test, three candles stand 1 m outside a hole, 20 cm apart, and the wall is 3 m inside. Where are the spots, and in what order?

Show solution
By straight lines through the hole, the spots lie on the wall at \(3\times20=60\) cm spacing, in reversed order: the left candle lights the right spot. The reversal is the inversion of the camera obscura image.
Lesson 05

How we see

For the general reader. Euclid and Ptolemy explained vision by rays going out from the eye. Aristotle and the atomists had the object send something to the eye, but without a geometry. Ibn al-Haytham joined the two: light from every point of the object enters the eye, and a geometrical rule picks out one ray from each point, so that the eye receives an ordered picture.

Goals
  • Follow the arguments against visual rays from the eye.
  • Understand the rule of perpendicular rays and the visual cone.
  • Use size and distance to analyse perception, including the Moon illusion and binocular vision.

Against rays from the eye

Strong light hurts the eye, and bright objects leave afterimages: both show that something acts on the eye from outside. When we open our eyes at night we see the stars at once; it is implausible that something issuing from the eye fills the space up to the stars in an instant. And if rays did go out, they would still need to bring something back for the eye to perceive. He concluded that the rays from the eye are superfluous.

The problem of confusion

If every point of an object sends light in all directions, every point of the eye’s surface receives light from every point of the object. Why then do we not see a blur? His answer used the geometry of the eye as Galen described it, a set of nested spherical surfaces sharing a centre.

Rule of perpendicular rays (Optics I)

Of all the rays that reach the front of the eye from a given point of the object, exactly one meets the cornea perpendicularly: the one directed at the common centre of the eye’s surfaces. It passes into the eye without refraction. Only these perpendicular rays produce distinct vision. Together they form a cone with vertex at the centre of the eye and the visible object as base, and points of the object correspond one to one, in the same order, to points on the sensitive surface of the crystalline humour.

Refracted oblique rays were real in his account and contributed weakly; Book VII uses them to explain why objects slightly outside the cone are seen. The theory gave Euclid’s visual cone a physical basis with the direction of travel reversed. Its weakness is that the image is formed in front of the centre and is never inverted; Kepler would locate the real image on the retina in 1604, inverted.

Perception and judgement

Book II argues that we perceive size by combining the angle an object subtends with a judgement of its distance, made quickly and without awareness. Historians compare this to Hermann von Helmholtz’s “unconscious inference”. Book III, and a separate letter, explain the Moon illusion by this route. The Moon near the horizon subtends the same angle as at the zenith, but the sky near the horizon, with intervening land, is judged farther away, so the Moon is judged larger. Cleomedes had given a related account in antiquity.

Size–distance relation

For an object of size \(S\) at distance \(D\), the visual angle is \(\theta=2\arctan\frac{S}{2D}\approx S/D\). If the perceived distance is \(D'\), the perceived size is \(S'\approx\theta D'\).

Binocular vision

Book III studies seeing with two eyes. When both eyes fix on a point, objects whose images fall on corresponding places in the two eyes are seen single; others are seen double. He tested this with a board marked with lines and small objects at different distances.

Exercise 5.1Two eyes are at \(E_1,E_2\) and fix a point \(F\). Show that every point \(P\) of the circle through \(E_1,E_2,F\), on the same arc as \(F\), subtends the same angle \(\angle E_1PE_2\) at the two eyes. Why does this make the circle a natural candidate for the set of points seen single (the horopter)?

Show solution
By the inscribed angle theorem, all points on one arc of the circle see the chord \(E_1E_2\) under the same angle. So the two eyes converge by the same amount for \(P\) as for \(F\), and the angles from each eye’s line of sight to \(P\) are equal and in the same sense. The images fall at corresponding places. This geometric horopter is now called the Vieth–Müller circle.

Exercise 5.2The Moon subtends about \(0.52^\circ\). An observer judges the horizon sky to be 2.5 times as far as the zenith sky. By the size–distance relation, how much larger should the horizon Moon look, and what does the angular size alone predict?

Show solution
\(S'\propto\theta D'\) with \(\theta\) unchanged, so the horizon Moon looks 2.5 times larger. The angle alone predicts no change, and in fact the horizon Moon is very slightly smaller, since it is about one Earth radius farther away. The illusion lies in judgement, as Ibn al-Haytham held, not in refraction, as Ptolemy had suggested in the Almagest.
Lesson 06

Reflection and mirrors

For the general reader. That a mirror returns light at the angle it receives it was known in antiquity. Ibn al-Haytham stated the law in full, for plane and curved mirrors, and built a brass instrument to test it. He also studied burning mirrors, which concentrate sunlight, and found that spherical ones cannot bring rays to a single point.

Goals
  • State the law of reflection in his complete form.
  • Prove Heron’s least-path property and locate images.
  • Compare the focusing of spherical and paraboloidal mirrors.
Law of reflection (Optics IV)

For a mirror of any shape, the incident ray, the reflected ray and the normal to the surface at the point of reflection lie in one plane, and the reflected ray makes the same angle with the normal as the incident ray, on the other side.

For curved mirrors the normal is the normal to the tangent plane: the radius for a sphere, the perpendicular to the axis through the surface point for a cylinder. His instrument was a graduated brass ring set vertically, with sights and interchangeable mirrors at the centre, so that a beam admitted through small holes could be traced after reflection.

Images

Book V adopts the rule, already used by Ptolemy, that the image of a point lies on the perpendicular from the point to the mirror (the cathetus, or for a sphere, the line through the centre), where the reflected ray, extended backward, meets it.

Proposition (Heron, Ibn al-Haytham)

For a plane mirror and points \(A,B\) on the same side, the path \(A\to P\to B\) with \(P\) on the mirror is shortest exactly when the angles of incidence and reflection are equal. The image \(A'\) of \(A\) is its mirror image in the plane, and all reflected rays from \(A\) appear to come from \(A'\).

Proof. Reflect \(A\) in the mirror to \(A'\). Then \(AP=A'P\), so \(AP+PB=A'P+PB\ge A'B\), with equality when \(P\) is on segment \(A'B\). At that point the angles are equal by vertical angles.

Burning mirrors

In On Paraboloidal Burning Mirrors he proved what earlier writers had asserted: a mirror shaped as a paraboloid of revolution reflects all rays parallel to its axis through one point, the focus. In On Spherical Burning Mirrors he showed that a concave spherical mirror sends parallel rays at different heights to different points on the axis, the effect now called spherical aberration. The Latin translation of the paraboloid treatise, probably by Gerard of Cremona, was read in medieval Europe.

Exercise 6.1Show that the parabola \(y^2=4fx\) reflects every ray parallel to the \(x\)-axis through the focus \((f,0)\).

Show solution
At \(P=(x_0,y_0)\) the tangent meets the \(x\)-axis at \(T=(-x_0,0)\). Then \(FT=x_0+f\), and \(FP=x_0+f\) by the focus–directrix property, so triangle \(FPT\) is isosceles and \(\angle FPT=\angle FTP\). The incoming ray is parallel to the axis, so it makes angle \(\angle FTP\) with the tangent. The reflected ray makes the same angle with the tangent on the other side, which is \(\angle FPT\): it passes through \(F\).

Exercise 6.2A concave spherical mirror of radius \(R\) receives a ray parallel to the axis at height \(h=R\sin\theta\). Show that the reflected ray crosses the axis at distance \(R/(2\cos\theta)\) from the centre. Evaluate for \(h\to0\) and \(h=R/2\).

Show solution
The normal at the point is the radius, at angle \(\theta\) to the axis; the ray reflects at angle \(\theta\) on the other side, making angle \(2\theta\) with the axis. The triangle formed by the centre \(C\), the point \(P\) and the crossing point \(X\) has angles \(\theta\) at \(C\) and at \(P\), so it is isosceles with \(CX=PX\); dropping the perpendicular from \(X\) to \(CP\) gives \(CX\cos\theta=R/2\). For small \(h\), \(CX\to R/2\), the paraxial focus. For \(h=R/2\), \(\theta=30^\circ\), \(CX=0.577R\): the outer rays cross closer to the mirror. No single focus exists.

Exercise 6.3Two plane mirrors meet at a right angle. Show that a ray in a plane perpendicular to both mirrors, after reflecting from each, returns parallel to its original direction.

Show solution
Take the mirrors along the axes. Reflection in the \(x\)-axis sends direction \((u,v)\) to \((u,-v)\); reflection in the \(y\)-axis then sends it to \((-u,-v)\). The final direction is the reverse of the first, so the ray returns parallel. Three mutually perpendicular mirrors do the same in space: the corner reflector.
Lesson 07

Alhazen’s problem

For the general reader. In a curved mirror, where must you look to see the reflection of a given object? For a plane mirror the answer takes one line. For a spherical mirror it leads to an equation of the fourth degree. Ibn al-Haytham solved it with conic sections, and the problem kept mathematicians busy for the next nine centuries.

Goals
  • State the problem and its plane reduction.
  • Derive the quartic equation and Huygens’s hyperbola.
  • Understand why straightedge and compass do not suffice.
Problem (Optics V)

Given a spherical mirror, convex or concave, and two points \(A\) (the eye) and \(B\) (the object), find the points \(P\) of the mirror at which light from \(B\) is reflected to \(A\).

By the law of reflection, \(A\), \(B\), \(P\) and the normal at \(P\) lie in one plane, and the normal passes through the centre \(O\). So \(P\) lies in the plane \(OAB\), and the problem is planar: given a circle and two points, find the points \(P\) of the circle where \(OP\) bisects the angle \(APB\). Ibn al-Haytham showed that there is one such point for a convex mirror and up to four for a concave one. He also treated cylindrical and conical mirrors.

His solution

He reduced the problem to a sequence of six lemmas. The key lemma asks for a line through a given point meeting a circle and a line so that the intercepted segment has a given length, a type of problem the Greeks called neusis. He solved it by intersecting the circle with a hyperbola, following Apollonius. The text has scribal errors and was long thought obscure; modern analysis, notably by Muṣṭafā Naẓīf and A. Mark Smith, shows the argument complete.

The modern equation

Put the circle as the unit circle in \(\mathbb C\), with \(A=a\), \(B=b\), and \(P=z\), \(|z|=1\). The condition that the radius bisects the angle between \(PA\) and \(PB\) is that \((a-z)/z\) and \((b-z)/z\) have arguments symmetric about 0, that is,

\[ \frac{(a-z)(b-z)}{z^2}\in\mathbb R. \]
Theorem (quartic form)

The reflection points are the roots on the unit circle of

\[ \bar a\bar b\,z^4-(\bar a+\bar b)\,z^3+(a+b)\,z-ab=0. \]

Equivalently, with \(ab=p+iq\) and \(a+b=s+it\), they are the intersections of the unit circle with the rectangular hyperbola

\[ q\,(x^2-y^2)-2p\,xy-t\,x+s\,y=0, \]

which passes through the centre \(O\).

Derivation. A complex number \(w\) is real iff \(w=\bar w\). On the unit circle \(\bar z=1/z\), so \(\overline{(a-z)(b-z)\bar z^2}=(\bar a-\bar z)(\bar b-\bar z)z^2=\bar a\bar bz^2-(\bar a+\bar b)z+1\). Setting \((a-z)(b-z)\bar z^2\) equal to this and multiplying by \(z^2\) gives the quartic. For the hyperbola, the imaginary part of \((a-z)(b-z)\bar z^2=ab\,\bar z^2-(a+b)\bar z+1\) must vanish (using \(z\bar z=1\)); expanding \(\bar z=x-iy\) gives the equation. Its quadratic part \(q(x^2-y^2)-2pxy\) has trace zero, so the hyperbola is rectangular. Huygens, between 1669 and 1673, also solved the problem with a single hyperbola valid in every case; the form above is a convenient modern version of such a solution.

Impossibility

In 1965 Jack M. Elkin showed, using the resolvent cubic of the quartic, that for some positions of \(A\) and \(B\) the reflection point is not constructible with straightedge and compass. Peter M. Neumann gave an independent proof in 1998. Ibn al-Haytham’s use of conics was therefore necessary, not a lack of ingenuity.

Exercise 7.1Let \(A\) and \(B\) be at the same distance from \(O\), symmetric about the \(x\)-axis: \(a=re^{i\varphi}\), \(b=re^{-i\varphi}\). Find two reflection points without solving the quartic, and check them against it.

Show solution
By symmetry, \(z=\pm1\) work: at \(z=1\) the radius is the \(x\)-axis, which bisects the angle between the symmetric rays. Check: \(ab=r^2\), \(a+b=2r\cos\varphi\), both real, so the quartic is \(r^2z^4-2r\cos\varphi\,z^3+2r\cos\varphi\,z-r^2=0\), and \(z=1\) gives \(r^2-2r\cos\varphi+2r\cos\varphi-r^2=0\). Likewise \(z=-1\). The quartic factors as \((z^2-1)(r^2z^2-2r\cos\varphi\,z+r^2)\). The second factor has roots with product 1; they are \(z=(\cos\varphi\pm i\sqrt{r^2-\cos^2\varphi})/r\), complex conjugates of modulus 1, when \(\cos^2\varphi<r^2\), and real roots off the circle otherwise. So there are four reflection points when \(|\cos\varphi|<r\) and two when \(|\cos\varphi|>r\).

Exercise 7.2For \(a=-0.45+0.30i\) and \(b=0.40-0.20i\), write the hyperbola of the theorem, and check that \(z=e^{i\cdot 47.04^\circ}\) is approximately on it.

Show solution
\(ab=-0.12+0.21i\), so \(p=-0.12,\ q=0.21\); \(a+b=-0.05+0.10i\), so \(s=-0.05,\ t=0.10\). The hyperbola is \(0.21(x^2-y^2)+0.24xy-0.10x-0.05y=0\). At \(x=\cos47.04^\circ=0.6814\), \(y=0.7319\): \(0.21(0.4643-0.5357)+0.24(0.4987)-0.0681-0.0366=-0.0150+0.1197-0.1047=0.0000\). This is one of the four points in the hero figure.

Exercise 7.3Let the eye look at itself: \(b=a=\rho u\) with \(|u|=1\), \(0<\rho<1\). Factor the quartic and find the reflection points. Interpret them.

Show solution
The quartic is \(\bar a^2z^4-2\bar az^3+2az-a^2=(\bar az^2-a)(\bar az^2+a)-2z(\bar az^2-a)=(\bar az^2-a)(\bar az^2-2z+a)\). The first factor gives \(z^2=a/\bar a=u^2\), so \(z=\pm u\). The second has roots \(z=(1\pm\sqrt{1-\rho^2})/\bar a\), of modulus \((1\pm\sqrt{1-\rho^2})/\rho\ne1\). So only two points: the ends of the diameter through \(A\), where the ray returns along the normal. A plane mirror would give one point; the extra roots of the circle problem have left the mirror.
Lesson 08

Refraction

For the general reader. Light bends when it passes from air into water or glass. Ptolemy had measured the bending. Ibn al-Haytham measured it again with a better instrument, stated rules that the measurements obey, and offered a mechanical picture: light moves more slowly in a denser medium, and its motion splits into parts along and across the surface. He did not find the exact law, but his picture became the starting point for René Descartes.

Goals
  • Know the refraction apparatus and the qualitative rules of Book VII.
  • Analyse the decomposition of motion and its consequences.
  • Derive the sine law from Fermat’s least time and compare with his “easier path”.

Apparatus and rules

The instrument of Book VII was a graduated brass disc, set vertically and half immersed in water (or with glass blocks at its centre), with sights to admit a narrow beam at a measured angle. Let \(i\) be the angle of incidence, \(r\) the angle of refraction, both from the normal, and \(d=i-r\) the deviation. He preferred to state results in terms of \(d\).

Rules of refraction (Optics VII, summarized)
  1. The incident ray, the refracted ray and the normal lie in one plane, the incident and refracted rays on opposite sides of the normal.
  2. A ray along the normal is not refracted.
  3. From a rarer to a denser medium the ray bends toward the normal; from denser to rarer, away from it, and the path is reversible.
  4. As \(i\) increases, \(r\) and \(d\) increase, and \(d\) is less than \(i\).
  5. For the same \(i\), the deviation is greater for a denser second medium (glass more than water).

He also stated rules about how the ratio of deviation to incidence changes with \(i\). Muṣṭafā Naẓīf showed that some of these hold only for the media and ranges of angle that were tested.

The mechanical picture

He compared light to a ball thrown at a thin board. A ball striking perpendicularly breaks through; a ball striking obliquely is deflected. He decomposed the motion of light into a component along the normal and a component parallel to the surface. A denser medium resists the motion more, so light is slower there. Motion along the normal, he argued, is the strongest and passes most easily, so a ray entering a denser medium turns toward the normal. He also remarked that refracted light takes a path that is “easier and quicker”.

Theorem (law of refraction; Ibn Sahl, Snell, Descartes)

For light passing between media in which it has speeds \(v_1\) and \(v_2\), Fermat’s principle of least time gives

\[ \frac{\sin i}{v_1}=\frac{\sin r}{v_2}, \qquad\text{that is}\qquad n_1\sin i=n_2\sin r, \quad n_k=c/v_k. \]

Proof. Let the light go from \(A=(0,h_1)\) to \(B=(L,-h_2)\), crossing the surface \(y=0\) at \((x,0)\). The time is \(T(x)=\sqrt{x^2+h_1^2}/v_1+\sqrt{(L-x)^2+h_2^2}/v_2\). Setting \(T'(x)=0\) gives \(x/(v_1\sqrt{x^2+h_1^2})=(L-x)/(v_2\sqrt{(L-x)^2+h_2^2})\), which is \(\sin i/v_1=\sin r/v_2\). So light bends toward the normal exactly when it is slower in the second medium, as Ibn al-Haytham held.

The sine law itself had been found around 984 by Ibn Sahl, in a treatise on burning lenses, though it did not enter Ibn al-Haytham’s optics. In Europe Willebrord Snell found it in 1621, and Descartes published it in 1637, deriving it from a decomposition of motion like Ibn al-Haytham’s.

Exercise 8.1Suppose light is slower in water, \(v_2<v_1\), and the component of velocity parallel to the surface is unchanged. Show that the ray bends away from the normal. What must happen to the parallel component for it to bend toward the normal, as observed?

Show solution
Parallel component \(v_1\sin i=v_2\sin r\) gives \(\sin r=(v_1/v_2)\sin i>\sin i\): away from the normal. To bend toward it with \(v_2<v_1\), the parallel component must decrease by a larger factor than the speed, as in Ibn al-Haytham’s account. Descartes kept the parallel component and assumed light faster in water, which gives the correct sine law but the wrong speeds; Fermat’s least time gives the correct law with light slower in water.

Exercise 8.2With \(n=1.333\) for water, compute \(r\) and \(d\) for \(i=30^\circ,60^\circ,80^\circ\). Check rule 4, and find how the ratio \(d/i\) changes.

Show solution
\(r=\arcsin(\sin i/1.333)\): \(22.0^\circ, 40.5^\circ, 47.6^\circ\), so \(d=8.0^\circ, 19.5^\circ, 32.4^\circ\). Both \(r\) and \(d\) increase and \(d<i\). The ratio \(d/i\) grows: 0.27, 0.33, 0.41. For glass, \(n=1.5\), \(i=80^\circ\) gives \(r=41.0^\circ\), \(d=39.0^\circ\), close to the limit.

Exercise 8.3Use the sine law to explain why there is a limit to the angle of refraction from air into water, and find it.

Show solution
Since \(\sin i\le1\), \(\sin r\le1/1.333=0.750\), so \(r\le48.6^\circ\). By reversibility, light going from water to air at more than \(48.6^\circ\) cannot emerge and is totally reflected.
Lesson 09

Doubts concerning Ptolemy

For the general reader. Ptolemy’s planetary models predicted well, but some of their circles moved in ways no real body could. Ibn al-Haytham listed these contradictions in a book of doubts. He did not reject Ptolemy’s astronomy; he demanded that its models describe physical bodies consistent with the principles Ptolemy himself accepted.

Goals
  • Know the three phases of his astronomy: physical spheres, doubts, new model.
  • Understand the objection to the equant quantitatively.
  • See how the critique set the programme of the Maragha astronomers.

The configuration of the world

An early treatise, On the Configuration of the World, presents Ptolemy’s models as solid, nested spheres in the manner of the Planetary Hypotheses, with the Earth stationary at the centre. It was translated into Spanish, Latin and Hebrew, and it influenced Georg Peurbach’s Theoricae novae planetarum in the fifteenth century.

The doubts

Al-Shukūk ʿalā Baṭlamyūs, written around 1025–28, examines the Almagest, the Planetary Hypotheses and the Optics. Its main objections concern the astronomy:

  • The equant. In the models for the planets the epicycle centre moves uniformly in angle about the equant point, not about the centre of the circle that carries it. A solid sphere cannot rotate uniformly about a point other than its centre.
  • The Moon. In Ptolemy’s lunar model the epicycle’s apogee is aligned toward a point other than the centre of its carrier, again a motion no sphere can produce.
  • Latitudes. The models for planetary latitudes require circles whose planes oscillate, and the Almagest and Planetary Hypotheses do not agree.

He argued that motions assigned to imagined lines and points cannot cause the motion of real bodies, and that Ptolemy had admitted departing from his own principles. The correct arrangement of bodies, he concluded, must exist and had not yet been found.

Proposition (non-uniformity about the centre)

Let a point move on a circle of radius 1 and centre \(C\) so that its direction from the equant \(Q\), at distance \(e\) from \(C\), turns uniformly at rate \(\omega\). Then its angular velocity about \(C\) is not constant: to first order in \(e\) it is \(\omega(1-e\cos M)\) at mean anomaly \(M\) measured from apogee, ranging from about \(\omega(1-e)\) at apogee to \(\omega(1+e)\) at perigee.

Proof. In Ptolemy’s arrangement the equant lies on the apogee side of \(C\). Put \(C=0\), \(Q=e\) on the real axis, apogee at angle 0, and the moving point at \(e^{i\theta}\). Since \(e^{i\theta}-e=e^{i\theta}(1-e\,e^{-i\theta})\), the condition \(\arg(e^{i\theta}-e)=M\) gives \(M=\theta+e\sin\theta+O(e^2)\), so \(\theta=M-e\sin M+O(e^2)\) and \(\dot\theta=\omega(1-e\cos M)+O(e^2)\). As a check, at apogee the point is at distance \(1-e\) from \(Q\), so uniform turning about \(Q\) needs linear speed \((1-e)\omega\) there, and \((1+e)\omega\) at perigee.

The model of the motions

Around 1038 he wrote The Model of the Motions of Each of the Seven Planets, which survives incomplete in a single damaged manuscript. Roshdi Rashed’s edition shows a geocentric kinematics without the equant, using geometry and trigonometry to describe the apparent motions. The Maragha astronomers of the thirteenth century, above all Nasir al-Din al-Tusi, took up the same programme and solved it with new devices.

The Milky Way

In a short work he used parallax to locate the Milky Way. A body near the Earth shifts against the stars as the observer’s position changes; the Milky Way shows no such shift, so it lies far beyond the Moon and is not a phenomenon of the upper air, as Aristotle had held.

Exercise 9.1Ptolemy’s Mars has \(e=6\) and deferent radius 60 (Lesson 08 of the Ptolemy course). By what ratio does the epicycle centre’s angular speed about the deferent centre differ between perigee and apogee?

Show solution
Here \(e=0.1\) relative to the radius, so the ratio is about \((1+e)/(1-e)=1.1/0.9\approx1.22\). A sphere carrying the epicycle would have to speed up and slow down by over 20 percent each revolution, which no uniformly rotating sphere does. That is the physical core of the objection.

Exercise 9.2The Moon is about 60 Earth radii away. Two observers 1,000 km apart sight it against the stars. Estimate the parallax shift, and explain why the absence of any detectable shift places the Milky Way much farther than the Moon.

Show solution
The baseline is \(1000/6371\approx0.157\) Earth radii, so the shift is about \(0.157/60\) rad \(\approx0.15^\circ\), about a third of the Moon’s diameter and easily seen. An object below the Moon would shift more. Since the Milky Way keeps its place among the stars, its distance is far greater than the Moon’s.
Lesson 10

Power sums, parallels, lunes, numbers

For the general reader. Ibn al-Haytham wrote about twenty-five mathematical works. Four topics stand out. He found the sum of fourth powers and used it to compute a volume that Archimedes had not. He tried to prove Euclid’s parallel postulate. He generalized the ancient theorem on lunes. And he solved a problem in congruences with what is now called Wilson’s theorem.

Goals
  • Derive the sum of fourth powers by his recursive identity.
  • Compute the volume of a paraboloid of the second kind as a limit of sums.
  • Follow his approach to parallels and its quadrilateral.
  • Solve his congruence problem.

Sums of powers

He justified the formulas by a geometric dissection of rectangles, which amounts to the identity below, valid for every \(k\ge1\).

Lemma (Ibn al-Haytham)

For integers \(n,k\ge1\), \[ (n+1)\sum_{j=1}^{n}j^k=\sum_{j=1}^{n}j^{k+1}+\sum_{p=1}^{n}\sum_{j=1}^{p}j^k. \]

Proof. \(\sum_{p=1}^{n}\sum_{j=1}^{p}j^k=\sum_{j=1}^{n}(n-j+1)\,j^k\). Adding \(\sum j^{k+1}\) gives \(\sum (n+1)j^k\).

Theorem (sum of fourth powers)

\[ \sum_{j=1}^{n}j^4=\Big(\frac n5+\frac15\Big)\,n\,\Big(n+\frac12\Big)\Big((n+1)n-\frac13\Big)=\frac{n(n+1)(2n+1)(3n^2+3n-1)}{30}. \]

The first form is close to the one he stated in words. The lemma with \(k=3\) and the known sum of cubes gives it.

The paraboloid

Archimedes had found the volume of a paraboloid formed by turning a parabolic segment about its axis. Ibn al-Haytham also turned it about the chord perpendicular to the axis, which gives a lemon-shaped solid. Slicing perpendicular to the chord gives discs whose radii are \(k(b^2-y^2)\), and the sum of their volumes needs fourth powers.

Theorem (Ibn al-Haytham)

Let the parabola \(x=ky^2\), \(|y|\le b\), with chord \(x=kb^2\), be rotated about that chord. The solid has volume \(\tfrac{8}{15}\) of the circumscribed cylinder, of radius \(kb^2\) and height \(2b\).

Proof by sums. Divide \([0,b]\) into \(n\) parts of width \(b/n\). The slice at \(y_j=jb/n\) has radius \(kb^2(1-j^2/n^2)\). Half the volume is approximated by \(\sum_j \pi k^2b^4(1-j^2/n^2)^2\,b/n=\frac{\pi k^2b^5}{n}\sum_j\big(1-\frac{2j^2}{n^2}+\frac{j^4}{n^4}\big)\). Using \(\sum j^2\sim n^3/3\) and \(\sum j^4\sim n^5/5\), the limit is \(\pi k^2b^5(1-\tfrac23+\tfrac15)=\tfrac{8}{15}\pi k^2b^5\), which is \(\tfrac8{15}\) of the half-cylinder \(\pi(kb^2)^2b\). He bounded the sums above and below and concluded by a double reductio, in the manner of Archimedes.

Parallels

In his commentary on the premises of Euclid’s Elements and a treatise on the parallel postulate, he defined parallel lines as equidistant ones, and argued that a perpendicular moving along a line, keeping at right angles, traces a straight line with its other end. The argument turns on quadrilaterals with right angles, of the kinds later studied by Giovanni Girolamo Saccheri and Johann Heinrich Lambert; for a quadrilateral with three right angles he tried to show that the fourth is also right. In hyperbolic geometry the fourth angle is acute, so the attempt needs an assumption equivalent to the postulate, here the motion argument. Omar Khayyam objected to the use of motion in geometry, and al-Tusi criticized the argument in turn. The three-right-angle figure is sometimes called the Ibn al-Haytham–Lambert quadrilateral.

Lunes and congruences

Hippocrates of Chios had squared certain lunes. Ibn al-Haytham proved a general version: for a right triangle inscribed in a semicircle, the two lunes on the legs together equal the triangle. In number theory he asked for a number that leaves remainder 1 when divided by 2, 3, 4, 5 and 6 and is divisible by 7, and he observed that for a prime \(p\), the number \((p-1)!+1\) is divisible by \(p\). The statement is Wilson’s theorem, named after an eighteenth-century rediscovery.

Exercise 10.1Check the sum of fourth powers for \(n=3\), and derive it from the lemma with \(k=3\), using \(\sum j^3=\big(\tfrac{n(n+1)}2\big)^2\).

Show solution
\(1+16+81=98\), and \(3\cdot4\cdot7\cdot35/30=98\). From the lemma, \(\sum j^4=(n+1)\sum j^3-\sum_{p}\sum_{j\le p}j^3=(n+1)\tfrac{n^2(n+1)^2}{4}-\sum_{p=1}^n\tfrac{p^2(p+1)^2}{4}\). Expanding \(p^2(p+1)^2=p^4+2p^3+p^2\) puts \(\tfrac14\sum p^4\) on the right; collecting, \(\tfrac54\sum p^4=\tfrac{n^2(n+1)^3}{4}-\tfrac12\sum p^3-\tfrac14\sum p^2\), and substituting the known sums gives the formula.

Exercise 10.2Verify the \(\tfrac8{15}\) ratio by integration, and compare with Archimedes’s solid (rotation about the axis), which is \(\tfrac12\) of its cylinder.

Show solution
\(V=\pi k^2\int_{-b}^{b}(b^2-y^2)^2dy=2\pi k^2\big(b^5-\tfrac23b^5+\tfrac15b^5\big)=\tfrac{16}{15}\pi k^2b^5\), and the cylinder is \(\pi k^2b^4\cdot2b=2\pi k^2b^5\). Ratio \(\tfrac8{15}\). For rotation about the axis, the radius at \(x\) is \(\sqrt{x/k}\), so the area of the slice is linear in \(x\) and only \(\sum j\) is needed: ratio \(\tfrac12\).

Exercise 10.3Find all numbers that leave remainder 1 on division by 2, 3, 4, 5, 6 and are divisible by 7. Show that \(6!+1\) is one of them.

Show solution
The condition is \(N\equiv1\pmod{60}\) and \(N\equiv0\pmod 7\). Write \(N=60m+1\); \(60\equiv4\pmod7\), so \(4m+1\equiv0\), \(m\equiv5\pmod7\). Hence \(N=301+420t\). \(6!+1=721=301+420\). It works because \(6!\) is divisible by 2,…,6 and \(6!\equiv-1\pmod7\) by Wilson’s theorem.

Exercise 10.4A right triangle with legs \(a,b\) and hypotenuse \(c\) is inscribed in a semicircle on \(c\). Semicircles are drawn outward on the legs. Show that the two lunes (the parts of the small semicircles outside the large one) have total area equal to the triangle.

Show solution
Area of lunes = (small semicircles) + (triangle) − (large semicircle) \(=\tfrac\pi8(a^2+b^2)+T-\tfrac\pi8c^2=T\), by Pythagoras.
Part III · Life and legacy
Lesson 11

The Cairo years and the last works

From the 1020s to his death Ibn al-Haytham seems to have lived in Cairo as a private scholar, copying, teaching and writing. The lists of works show a steady output. Between February 1027 and July 1028 alone he added 21 titles. By October 1038 the total was 92. Many works answer questions sent by others, such as the replies to the scholar Ibn Maʿdān on Ptolemy’s account of the apparent enlargement of stars near the horizon.

Among the shorter optical works of these years are On the Light of the Moon, which argues that the Moon shines with borrowed light emitted from every point of its surface, as a self-luminous body would; On the Halo and the Rainbow (1028), whose explanation fails but which Kamāl al-Dīn al-Fārisī later revised; On the Light of the Stars, which holds that the stars and planets other than the Moon shine by their own light; On Shadows, which distinguishes full shadow and penumbra; and the Discourse on Light, a summary of his doctrine. In mathematics he wrote Analysis and Synthesis, on the method of discovery in geometry, arithmetic and astronomy, and On Known Things, on what it means for a quantity or figure to be given.

Biographers mention students, including a certain Surkhāb of Semnan and the Egyptian amīr Abū al-Wafāʾ al-Mubashshir ibn Fātik; the details are thin. He died in Cairo around 1040, or soon after. No grave is known.

Exercise 11.1Ibn al-Haytham explained the Moon’s light as re-emitted from every point of its surface, not mirrored. Suppose the Moon were a perfectly polished sphere. Using the law of reflection, describe what an observer on Earth would see at full Moon.

Show solution
A mirror sphere shows the Sun as a tiny virtual image near the point where the normal bisects the Sun–Moon–Earth angle, so at full Moon a single bright point near the disc centre, with the rest dark. The real Moon is lit across its whole visible disc, so its surface must scatter light from every point, as he argued.
Lesson 12

From Alhacen to Kepler

In the Islamic world

The Optics was little read in Arabic after the eleventh century, until Kamāl al-Dīn al-Fārisī wrote his commentary, the Tanqīḥ al-Manāẓir (Revision of the Optics), early in the fourteenth century, at the suggestion of his teacher Quṭb al-Dīn al-Shīrāzī. Using a glass sphere filled with water as a model raindrop, al-Fārisī explained the primary and secondary rainbows by two refractions and one or two reflections, at about the same time as Theodoric of Freiberg in Europe. The Doubts set the agenda for the astronomers of Maragha. Al-Tusi criticized Ibn al-Haytham’s treatment of parallels and wrote his own.

In Latin Europe

The Latin De aspectibus, made around 1200 by an unknown translator, survives in about nineteen manuscripts, all missing the opening chapters of Book I. Roger Bacon, John Pecham and Witelo built on it the medieval science of perspectiva, which became a university subject. Lorenzo Ghiberti drew on it in his notes on art. Risner’s edition of 1572 put it in print next to Witelo. Johannes Kepler’s Ad Vitellionem paralipomena (1604) started from this tradition and replaced the rule of perpendicular rays with the retinal image. Snell, Descartes, Pierre de Fermat and Huygens all read him.

Assessment

  • Optics: intromission given a geometry; light and vision separated into physics, physiology and psychology; the law of reflection in full; the first systematic use of the dark chamber.
  • Method: repeated tests with instruments, stated beforehand, and results reported against claims. Historians differ over how far to call this “the scientific method”; Gerald Toomer warned against isolating him from his predecessors. The Optics is, in any case, an early sustained example of physics argued from controlled experiment.
  • Mathematics: Alhazen’s problem; power sums and the paraboloid; the quadrilateral of the parallel theory.
  • Astronomy: the demand that models be physically realizable, which led to the equant-free models of Maragha and, through channels still debated, to Copernicus.
  • Honours: the lunar crater Alhazen and asteroid 59239 Alhazen; his portrait on the Iraqi 10,000-dinar note (2003); the theme of UNESCO’s International Year of Light in 2015.

Final projectBuild a simple dark chamber with a shutter and a card, or a box camera. Repeat the candle test of Lesson 04 and photograph the Sun’s image. Then write two pages comparing your procedure with his, and saying which of his principles your observations test and which they assume.

Reference

Sources and further reading

Biography and context

  1. A. I. Sabra, Ibn al-Haytham, Dictionary of Scientific Biography, vol. 6, Scribner, 1972.
  2. J. J. O’Connor and E. F. Robertson, Abu Ali al-Hasan ibn al-Haytham, MacTutor History of Mathematics.
  3. Ibn al-Haytham, Wikipedia (with extensive references).
  4. A. I. Sabra, One Ibn al-Haytham or two? An exercise in reading the bio-bibliographical sources, Zeitschrift für Geschichte der arabisch-islamischen Wissenschaften 12 (1998); and the sequel in vol. 15 (2002–03).
  5. D. C. Lindberg, Theories of Vision from al-Kindi to Kepler, University of Chicago Press, 1976.

Works and mathematics

  1. A. I. Sabra, The Optics of Ibn al-Haytham, Books I–III: On Direct Vision, translation and commentary, Warburg Institute, 1989.
  2. A. M. Smith, Alhacen’s Theory of Visual Perception (2001), Alhacen on the Principles of Reflection (2006), Alhacen on Image-Formation and Distortion in Mirrors (2008), Alhacen on Refraction (2010), American Philosophical Society.
  3. Book of Optics and Alhazen’s problem, Wikipedia.
  4. Ibn al-Haytham, The Doubts concerning Ptolemy, ed. A. I. Sabra and N. Shehaby, Cairo, 1971.
  5. R. Rashed, Les mathématiques infinitésimales du IXe au XIe siècle, vols. 2–4, Al-Furqan, 1993–2002.
  6. J. M. Elkin, A deceptively easy problem, Mathematics Teacher 58 (1965).
  7. P. M. Neumann, Reflections on reflection in a spherical mirror, American Mathematical Monthly 105 (1998).
  8. V. J. Katz, Ideas of calculus in Islam and India, Mathematics Magazine 68 (1995).
  9. R. Rashed, The celestial kinematics of Ibn al-Haytham, Arabic Sciences and Philosophy 17 (2007).