By Alhazen (lbn al-Haytham), 965-1039 ; A. Mark Smith (editor, translation, commentary)
Read Online or Download Alhacen on the principles of reflection. A Critical Edition, with English Translation and Commentary, of Books 4 and 5 of Alhacen’s De aspectibus. Volume One - Introduction and Latin Text ; Volume two - English Translation PDF
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Extra resources for Alhacen on the principles of reflection. A Critical Edition, with English Translation and Commentary, of Books 4 and 5 of Alhacen’s De aspectibus. Volume One - Introduction and Latin Text ; Volume two - English Translation
In this case, reflection can occur from B to A, or from B’ to A from arc KDL facing angle KGL, or from arc LD’K facing angle LGK. Alhacen deals with both these cases in proposition 34, pp. 450-451. For the situation in which B, B’, and A lie on the same normal, as represented in the left-hand diagram of figure 10, he demonstrates that within the plane of the circle (which constitutes a great circle on the sphere from which the mirror is formed) reflection can occur from B or B’ to A at two corresponding points D and D’ on the mirror within respective arcs of the circle.
452-454 (see pp. lvi-lvii below for a detailed analysis), the analysis of reflection for the first case is complete. The remaining analysis of concave spherical mirrors centers on the case in which A and B lie on separate normals. Accordingly, in proposition 37, pp. 454-458, Alhacen shows, first, that if A and B lie outside the mirror on different normals, only one reflection can occur within the given plane. On the other hand, he continues, if those points lie inside the mirror, and if they are equidistant from the mirror’s centerpoint, there can be as few as two or as many as four (but not three) reflections, depending on whether the circle passing through the center of sight, the object-point, and the center of the mirror intersects arc KL.
Consequently, DN1 bisects angle BDV1. Earlier we established that BE1:E1A1 = BG:GA1, so, if we draw a line from point E1 parallel to A1S1, it will pass through point Q1, cutting lines A1B and S1B equiproportionally. Hence, BE1:E1A1 = BQ1:Q1S1, from which it follows that BQ1:Q1S1 = BG:GA1. But Q1 is where DM1, which bisects angle BDS1, cuts line BS1. Therefore, given the similarity of triangles BM1Q1 and Q1DS1, it follows that corresponding sides are proportional. , BQ1:Q1S1 = M1Q1:Q1D. But BQ1:Q1S1 = BE1:E1A1 = BG:GA1, so M1Q1:Q1D = BG:GA1.