GENERAL ARTICLE The Easiest Proof of Fermat’s Principle Hasi Ray and Sudipto Roy This article presents the easiest direct proof of Fermat’s prin- ciple with the help of differential calculus for all types of sur- faces. This article is a contribution by Late Professor Hasi Ray, Study Center JnaganSiksha, 1. Introduction Kolkata and Sudipto Roy, David Hare School, Kolkata. Teaching and learning Fermat’s principle in optics often suffers from lack of satisfactory proof [1–5]. Students generally have the knowledge of the laws of reflection and refraction of light rays in geometrical optics, and they are also familiar with differen- tial calculus at the undergraduate level when they learn Fermat’s principle. However, the difficulties in teaching and grasping the concept motivated us to find the present logical proofs of Fermat’s principle of stationary optical paths. Fermat’s principle states that when a light ray moves from one fixed point to another fixed point, through any number of reflec- tions or refractions, the total optical path followed by the light ray should be stationary; it will either be minimum or maximum. For reflection and refraction at plane surfaces, the total optical path followed by the light ray should be a minimum, while for reflec- tion and refraction at curved surfaces, the total optical path fol- lowed by the light ray should be a maximum. It is also described as the ‘principle of least time’ in many books. The optical path is defined as the actual path followed by the light ray in vacuum or in air medium; so it is the product of the path length in the medium with the refractive index of the medium. The methodology we discuss in this article in the context of the most logical proof of Fermat’s principle, might be a convincing Keywords teaching and learning method. It might help in proper under- Optical path, stationary, reflection, standing of the concept and instill confidence in students to learn refraction, curved surface. physics. It will also enhance their inventing power by following RESONANCE | August 2018 861 GENERAL ARTICLE Figure 1. Reflection at a plane surface. easy tricks to overcome many difficult steps and the approxima- tions used to extract proper knowledge. The new method was applied by one of the authors (HR) in a BSc first year class, and received encouraging results. 2. Methodology I. Reflection and Refraction at Plane Surfaces Figures 1 and 2 are pictorially presents the concepts of reflection and refraction at plane surfaces respectively. In Figure 1, the lines AN and NB with arrow signs indicate the incident and the reflected rays and OO‘ is the plane of reflection with normal NC at the point of incidence N. The angle ∠ ANC=∠i (say) is the angle of incidence and the angle ∠ BNC= ∠r (say) is the angle of reflection. Say μ1 is the refractive index of the medium. The optical path followed by light ray to reach from the fixed point A to the fixed point B through the point of incidence Nisy = μ1.AN + μ1.NB. Perpendiculars are drawn on the line OO‘ from the fixed points A and B as lines AE and BF. Then we get two right triangles ΔANE and ΔBNF where ∠ AEN and ∠ BFN are the right angles in first and second triangles respectively. The height of the perpendiculars AE and BF (say h1 and h2)are constants since the points A and B are fixed. Again the distance between the feet of the perpendiculars EF = d is a constant. 862 RESONANCE | August 2018 GENERAL ARTICLE Figure 2. Refraction at a plane surface. Similarly, in Figure 2, the lines AN and NB with arrow signs indicate the incident and the refracted rays and OO‘ is the plane of refraction with normal CNM at the point of incidence N. The angle ∠ ANC= ∠i (say) is the angle of incidence and the angle ∠ BNM = ∠r (say) is the angle of refraction. Say μ1 and μ2 are the refractive indices of the first and the second media respectively – these are constant quantities. The optical path followed by light ray to reach from the fixed point A to the fixed point B through the point of incidence N is y = μ1. Perpendiculars are drawn on OO‘ from the fixed points A and B as lines AE and BF. Then we get two right triangles ΔANE and ΔBNF with ∠ AEN and ∠ BFN as the right angles in first and second triangles respectively. The height of the perpendiculars AE and BF (say h1 and h2)are constants since the points A and B are fixed. Again, the distance between the feet of the perpendiculars EF = d is a constant. In both the cases and in both the figures, the total optical path length will vary if the point of incidence N changes. Let us say the distance EN = x, so that FN = d − x. Accordingly, following the Pythagorean theorem for right triangles, the total optical path in case of reflection can be written as: RESONANCE | August 2018 863 GENERAL ARTICLE = μ . +μ . = μ 2 + 2 +μ { 2 + − 2} = y 1 AN 1 NB 1 (h1 x ) 1 h2 (d x) f1(x) (1) Similarly, in case of refraction the total optical path can be ex- pressed as: = μ . +μ . = μ 2 + 2 +μ { 2 + − 2} = y 1 AN 2 NB 1 (h1 x ) 2 h2 (d x) f2(x) (2) So in both the cases, we are able to write the total optical path y as a function of a variable x and it is differentiable with respect to the variable x. In case of reflection, after differentiation, (1) becomes: dy x (d − x) = μ1 − μ1 dx 2 + 2 { 2 + − 2} (h1 x ) h2 (d x) EN FN = μ − μ = μ cos ∠ANE − μ cos ∠BNF) 1 AN 1 BN 1 1 π π = μ cos − ∠i − μ cos − ∠r 1 2 1 2 = μ1 sin ∠i − μ1 sin ∠r (3) We know from the law of reflection that the angles ∠i = ∠r.Ap- dy = ⇒ = = plying it in (3) gives dx 0 y f1(x) stationary. Hence, Fermat’s principle is proved for reflection on a plane sur- face. In a reverse manner, if we consider that the Fermat’s principle is true, then we can derive the laws of reflection from (3) applying the derivative equal to zero. In case of refraction, after differentiation (2) becomes : 864 RESONANCE | August 2018 GENERAL ARTICLE d x (d − x) dx = μ1 − μ2 y 2 = 2 { 2 + − 2} (h1 x ) h2 (d x) EN FN = μ1 − μ2 = μ1 cos ∠ANE − μ2 cos ∠BNF) AN BN π π = μ cos − ∠i − μ cos − ∠r 1 2 2 2 = sin ∠i − μ2 sin ∠r (4) According to the law of refraction we know that μ1 sin ∠i = μ2 sin ∠r . dy = ⇒ = = Applying it in (4) gives dx 0 y f2(x) stationary. Hence, Fermat’s principle is proved for refraction on a plane sur- face. In reverse manner, if we consider that the Fermat’s principle is true, then we can derive the laws of refraction from (4) applying the derivative equal to zero. II. Reflection at Curved Surfaces – Convex and Concave Sur- faces. The reflections at the curved surfaces are pictorially presented in Figure 3a and in Figure 3b for convex and concave surfaces respectively. Let OO is the curved surface, and C is the center of curvature with R as the radius of curvature in both the figures. Let A and B be two fixed points situated in a medium of refractive index μ1 and B in a medium of refractive index μ2. The lines AN and NB with arrow signs indicate the incident and the reflected rays with N as the point of incidence or the point of reflection. Draw the normal CN at the point of incidence N on the curved surface by joining C with N so that CN = R. The total optical path followed by the light ray to reach from the fixed point A to the fixed point B through the point of incidence at N is y = μ1.AN+μ1.NB. The total path will vary if the point of incidence N changes. Perpendiculars are drawn on CN from the fixed points A and B as lines AE and BF; then we get two right triangles ΔANE and ΔBNF where ∠AEN and ∠BFN are the right angles in first and second triangles respectively in both the figures. The angle RESONANCE | August 2018 865 GENERAL ARTICLE Figure 3. (a) Reflection at a convex surface. (b) Reflec- (a) (b) tion at a concave surface. ∠ANE = ∠i (say) is the angle of incidence and the angle ∠BNF = ∠r (say) is the angle of reflection. Join A and B with C, then we get another two right triangles ΔAEC and ΔBFC when AC=h1 and BC=h2 are fixed distances i.e., constants. The angle ∠ACB = Θ is a constant angle. But the angle ∠ACN = θ varies if the point of incidence N changes. Applying the laws of triangles in ΔANC and ΔBNC, we can write: = μ . + μ . y 1 AN 1 NB = μ 2 + 2 − θ + μ { 2 + 2 − Θ − θ} 1 (h1 R 2h1R cos ) 1 h2 R 2h2R cos ( + f1(θ)(5) Differentiating with respect to θ, one can express the above ex- pression as: dy h R sin θ h R sin (Θ − θ) = μ 1 − μ 2 θ 1 1 d 2 + 2 − θ { 2 + 2 − Θ − θ } (h1 R 2h1R cos h2 R 2h2R cos( ) AE BF = μ R − μ R = μ R sin ∠ANE − μ R sin ∠BNF 1 AN 1 BN 1 1 = μ1R sin ∠i − μ1R sin ∠r = μ1R(sin ∠i − sin ∠r)(6) 866 RESONANCE | August 2018 GENERAL ARTICLE Figure 4.