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Refraction Problems in Geometric Optics

Part of the Lecture Notes in Mathematics book series (LNMCIME,volume 2087)

Abstract

We present a description of the far field and the near field problems for refraction when the source of energy is located at one point. The far field problem is solved using mass transportation and also a variant of the Minkowski method. Maxwell equations are developed and the boundary conditions studied to obtain Fresnel formulas. These are used to present a model for refraction that takes into consideration the energy used in internal reflection.

Keywords

  • Weak Solution
  • Wave Front
  • Radon Measure
  • Physical Constraint
  • Field Case

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Fig. 1
Fig. 2
Fig. 3
Fig. 4
Fig. 5
Fig. 6

Notes

  1. 1.

    Since the refraction angle depends on the frequency of the radiation, we assume radiation is monochromatic.

  2. 2.

    If θ 1 > θ c , then the phenomenon of total internal reflection occurs, see Fig. 1c.

  3. 3.

    For values of μ for different substances see http://en.wikipedia.org/wiki/Permeability_(electromagnetism)#Values_for_some_common_materials.

  4. 4.

    For relative permittivity of some substances see http://en.wikipedia.org/wiki/Relative_permittivity.

  5. 5.

    The relative permittivity is ε ∕ ε 0 and the relative permeability is μ ∕ μ 0; the index of refraction is defined by \(n = \sqrt{\epsilon _{r } \mu _{r}}\). The velocity of propagation \(v = \dfrac{1} {\sqrt{\epsilon \mu }}\). Since \(c = \dfrac{1} {\sqrt{\epsilon _{0 } \mu _{0}}}\), we get that n = c ∕ v.

  6. 6.

    Notice that from (28), the value of the field \({\mathbf{E}}^{i}\) at P is \({\mathbf{E}}^{i}(0,t) = \mathbf{E}_{0}^{i}\,\cos \left (\omega \,t\right )\). Hence we actually get \({J}^{i} = \dfrac{n_{1}} {4\pi } \vert \mathbf{E}_{0}^{i}{\vert }^{2}\cos \theta {_{i}\,\cos }^{2}(\omega \,t)\). Similarly, from (29), the value of the field \({\mathbf{E}}^{t}\) at P is \({\mathbf{E}}^{t}(0,t) = \mathbf{E}_{0}^{t}\,\cos \left (\omega \,t\right )\), and from (30), the value of the field \({\mathbf{E}}^{r}\) at P is \({\mathbf{E}}^{r}(0,t) = \mathbf{E}_{0}^{r}\,\cos \left (\omega \,t\right )\). Therefore we have \({J}^{r} = \dfrac{n_{1}} {4\pi } \vert \mathbf{E}_{0}^{r}{\vert }^{2}\cos \theta {_{i}\,\cos }^{2}(\omega \,t)\), and \({J}^{t} = \dfrac{n_{2}} {4\pi } \vert \mathbf{E}_{0}^{t}{\vert }^{2}\cos \theta {_{t}\,\cos }^{2}(\omega \,t)\). So we get formulas (33) because the factor \({\cos }^{2}(\omega t)\) cancels out.

  7. 7.

    We recall that the physical constraint for refraction is that xm ≥ κ for κ < 1, see Lemma 2.1.

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Acknowledgements

This material is based upon work partially supported by the National Science Foundation under Grants DMS-0901430 and DMS-1201401.

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Correspondence to Cristian E. Gutiérrez .

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Gutiérrez, C.E. (2014). Refraction Problems in Geometric Optics. In: Fully Nonlinear PDEs in Real and Complex Geometry and Optics. Lecture Notes in Mathematics(), vol 2087. Springer, Cham. https://doi.org/10.1007/978-3-319-00942-1_3

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