SNELL S LAW AND UNIFORM REFRACTION. Contents

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1 SNELL S LAW AND UNIFORM REFRACTION CRISTIAN E. GUTIÉRREZ Contents 1. Snell s law of refraction In vector form κ < κ > κ = Uniform refraction Surfaces with the uniform refracting propert: far field case κ = κ < κ > Uniform refraction: near field case 7 1. Snell s law of refraction 1.1. In vector form. Suppose Γ is a surface in R 3 that separates two media I and II that are homogeneous and isotropic. Let v 1 and v 2 e the velocities of propagation of light in the media I and II respectivel. The index of refraction of the medium I is definition n 1 = c/v 1, where c is the velocit of propagation of light in the vacuum, and similarl n 2 = c/v 2. If a ra of light having direction x S 2 and. June 15, 2011 Since the refraction angle depends on the frequenc of the radiation, we assume our light ra is monochromatic. 1

2 2 C. E. GUTIÉRREZ traveling through the medium I hits Γ at the point P, then this ra is refracted in the direction m S 2 through the medium II according with the Snell law in vector form: 1.1) n 1 x ν) = n 2 m ν), where ν is the unit normal to the surface to Γ at P going towards the medium II. This has several consequences: a) the vectors x, m, ν are all on the same plane called plane of incidence); ) the well known Snell law in scalar form n 1 sin θ 1 = n 2 sin θ 2, where θ 1 is the angle etween x and ν the angle of incidence), θ 2 the angle etween m and ν the angle of refraction), From 1.1), n 1 x n 2 m) ν = 0, which means that the vector n 1 x n 2 m is parallel to the normal vector ν. If we set κ = n 2 /n 1, then 1.2) x κ m = λν, for some λ R. Taking dot products we get λ = cos θ 1 κ cos θ 2, cos θ 1 = x ν > 0, and cos θ 2 = m ν = 1 κ 2 [1 x ν) 2 ], so 1.3) λ = x ν κ 1 κ 2 1 x ν) 2. Refraction ehaves differentl for κ < 1 and for κ > κ < 1. This means v 1 < v 2, and so waves propagate in medium II faster than in medium I, or equivalentl, medium I is denser than medium II. In this case the refracted ras tend to ent awa from the normal, that is the case for example, when medium I is glass and medium II is air. Indeed, from the scalar Snell law, sin θ 1 = κ sin θ 2 < sin θ 2 and so θ 1 < θ 2. For this reason, the maximum angle of refraction θ 2 is π/2 which, from the Snell law in scalar form, is achieved when sin θ 1 = n 2 /n 1 = κ. So there cannot e refraction when the incidence angle θ 1 is

3 SNELL S LAW 3 eond this critical value, that is, we must have 0 θ 1 θ c = arcsin κ. again from the Snell law in scalar form, Once 1.4) θ 2 θ 1 = arcsinκ 1 sin θ 1 ) θ 1 and it is eas to verif that this quantit is strictl increasing for θ 1 [0, θ c ], and therefore 0 θ 2 θ 1 π 2 θ c. We then have x m = cosθ 2 θ 1 ) cosπ/2 θ c ) = κ, and therefore otain the following phsical constraint for refraction: if κ = n 2 /n 1 < 1 and a ra of direction x through medium I 1.5) is refracted into medium II in the direction m, then m x κ. Notice also that in this case λ > 0 in 1.3). Conversel, given x, m S 2 with x m κ and κ < 1, it follows from 1.4) that there exists a hperplane refracting an ra through medium I with direction x into a ra of direction m in medium II κ > 1. In this case, waves propagate in medium I faster than in medium II, and the refracted ras tend to ent towards the normal. B the Snell law, the maximum angle of refraction denoted θ c is achieved when θ 1 = π/2, and θ c = arcsin1/κ). Once again from the Snell law in scalar form 1.6) θ 1 θ 2 = arcsinκ sin θ 2 ) θ 2 which is strictl increasing for θ 2 [0, θ c], and 0 θ 1 θ 2 π 2 θ c. We therefore otain the following phsical constraint for the case κ > 1: if a ra with direction x traveling through medium I 1.7) is refracted into a ra in medium II with direction m, then m x 1/κ. Notice also that in this case λ < 0 in 1.3). If θ 1 > θ c, then the phenomenon of total internal reflection occurs.

4 4 C. E. GUTIÉRREZ On the other hand, 1.6), if x, m S 2 with x m 1/κ and κ > 1, then there exists a hperplane refracting an ra of direction x through medium I into a ra with direction m in medium II. We summarize the aove discussion on the phsical constraints of refraction in the following lemma. Lemma 1.1. Let n 1 and n 2 e the indices of refraction of two media I and II, respectivel, and κ = n 2 /n 1. Then a light ra in medium I with direction x S 2 is refracted some surface into a light ra with direction m S 2 in medium II if and onl if m x κ, when κ < 1; and if and onl if m x 1/κ, when κ > κ = 1. This corresponds to reflection. It means 1.8) x m = λ ν. Taking dot products with x and then with m ields 1 m x = λ x ν and x m 1 = λ m ν, then x ν = m ν. Also taking dot product with x in 1.8) then ields λ = 2 x ν. Therefore m = x 2x ν)ν. 2. Uniform refraction 2.1. Surfaces with the uniform refracting propert: far field case. Let m S 2 e fixed, and we ask the following: if ras of light emanate from the origin inside medium I, what is the surface Γ, interface of the media I and II, that refracts all these ras into ras parallel to m? Suppose Γ is parameterized the polar representation ρx)x where ρ > 0 and x S 2. Consider a curve on Γ given rt) = ρxt)) xt) for xt) S 2. According to 1.2), the tangent vector r t) to Γ satisfies r t) xt) κ m) = 0. That is, [ρxt))] xt) + ρxt))x t) ) xt) κ m) = 0, which ields ρxt))1 κm xt)) ) = 0. Therefore 2.9) ρx) = 1 κ m x

5 SNELL S LAW 5 for x S 2 and for some R. To understand the surface given 2.9), we distinguish two cases κ < 1 and κ > κ = 1. When κ = 1 we see this is a paraoloid. Indeed, let m = e n, then a point X = ρx)x is on the surface 2.9) if X = x n. The distance from X to the plane x n = is x n, and the distance from X to 0 is X. So this is a paraoloid with focus at 0, directrix plane x n = and axis in the direction e n κ < 1. For > 0, we will see that the surface Γ given 2.9) is an ellipsoid of revolution aout the axis of direction m. Suppose for simplicit that m = e n, the nth-coordinate vector. If =, n ) R n is a point on Γ, then = ρx)x with x = /. From 2.9), κ n =, that is, n = κ n + ) 2 which ields κ 2 ) 2 n 2κ n = 2. This surface Γ can e written in the form 2.10) 2 n κ ) 2 1 κ 2 ) 2 + ) 2 = 1 1 κ 2 1 κ 2 which is an ellipsoid of revolution aout the n axis with foci 0, 0) and 0, 2κ/1 κ 2 )). Since = κ n + and the phsical constraint for refraction 1.5), e n κ is equivalent to n κ. That is, for refraction to occur must e in the upper 1 κ2 part of the ellipsoid 2.10); we denote this semi-ellipsoid Ee n, ). To verif that Ee n, ) has the uniform refracting propert, that is, it refracts an ra emanating from the origin in the direction ) e n, we check that 1.2) holds ) at each point. Indeed, if Ee n, ), then κe n 1 κ > 0, and κe n e n 0, and so κe n is an outward normal to Ee n, ) at. Rotating the coordinates, it is eas to see that the surface given 2.9) with κ < 1 and > 0 is an ellipsoid of revolution aout the axis of direction m with foci 0 and 2κ m. Moreover, the semi-ellipsoid Em, ) given 1 κ2 2.11) Em, ) = {ρx)x : ρx) = 1 κ m x, x Sn 1, x m κ},

6 6 C. E. GUTIÉRREZ has the uniform refracting propert, an ra emanating from the origin O is refracted in the direction m κ > 1. Due to the phsical constraint of refraction 1.7), we must have < 0 in 2.9). Define for > ) Hm, ) = {ρx)x : ρx) = κ m x 1, x Sn 1, x m 1/κ}. We claim that Hm, ) is the sheet with opening in direction m of a hperoloid of revolution of two sheets aout the axis of direction m. To prove the claim, set for simplicit m = e n. If =, n ) He n, ), then = ρx)x with x = /. From 2.12), κ n =, and therefore n = κ n ) 2 which ields 2 κ 2 1) n κ ) 2 ) 2 κ κ 2 1 κ 2 1 = 2. Thus, an point on He n, ) satisfies the equation 2.13) n κ ) 2 κ ) 2 ) 2 = 1 κ 2 1 κ2 1 which represents a hperoloid of revolution of two sheets aout the n axis with foci 0, 0) and 0, 2κ/κ 2 1)). Moreover, the upper sheet of this hperoloid of revolution is given n = κ κ κ / κ2 1 ) 2 and satisfies κ n > 0, and hence has polar equation ρx) = κ e n x 1. Similarl, the lower sheet satisfies κ n < 0 and has polar equation ρx) = κ e n x + 1. For a general m, a rotation, we otain that Hm, ) is the sheet with opening in direction m of a hperoloid of revolution of two sheets aout the axis of direction m with foci 0, 0) and 2κ κ 2 1 m.

7 SNELL S LAW 7 Notice that the focus 0, 0) is outside the region enclosed Hm, ) and the 2κ focus m is inside that region. The vector κm is an inward normal to κ 2 1 Hm, ) at, ecause 2.12) κm ) ) 2κ κ 2 1 m 2κ2 κ 2 1 = 2κ κ + 1 Clearl, κm ) m κ 1 and κm ) the uniform refraction propert. 2κ κm + κ 2 1 = κ 1) κ + 1 > 0. > 0. Therefore, Hm, ) satisfies We remark that one has to use H e n, ) to uniforml refract in the direction e n, and due to the phsical constraint 1.7), the lower sheet of the hperoloid of equation 2.13) cannot refract in the direction e n. From the aove discussion, we have proved the following. Lemma 2.1. Let n 1 and n 2 e the indexes of refraction of two media I and II, respectivel, and κ = n 2 /n 1. Assume that the origin O is inside medium I, and Em, ), Hm, ) are defined 2.11) and 2.12), respectivel. We have: i) If κ < 1 and Em, ) is the interface of media I and II, then Em, ) refracts all ras emitted from O into ras in medium II with direction m. ii) If κ > 1 and Hm, ) separates media I and II, then Hm, ) refracts all ras emitted from O into ras in medium II with direction m. 3. Uniform refraction: near field case The question we ask is: given a point O inside medium I and a point P inside medium II, find an interface surface S etween media I and II that refracts all ras emanating from the point O into the point P. Suppose O is the origin, and let Xt) e a curve on S. B the Snell law of refraction the tangent vector X t) satisfies X t) Xt) Xt) κ P Xt) ) = 0. P Xt)

8 8 C. E. GUTIÉRREZ That is, Xt) + κ P Xt) = 0. Therefore S is the Cartesian oval 3.14) X + κ X P =. Since f X) = X + κ X P is a convex function, the oval is a convex set. Department of Mathematics, Temple Universit, Philadelphia, PA address: gutierre@temple.edu