EE 543 Theory and Principles of Remote Sensing

1. EE 543Theory and Principles of Remote Sensing Reflection and Refraction from a Planar Interface

2. Outline • Reflection and Transmission at a planar interface • Boundary conditions • Fresnel reflection coefficients • Special Cases: Total reflection (critical angle), Total transmission (Brewster angle) • Rough Surface • Scattering from objects O. Kilic EE 543

3. Reflection and Transmission • When we consider interaction of em waves with a target, we must account for the effects of boundaries between media. • These boundary effects give rise to changes in the amplitude, phase and direction of propagation waves. • These in turn either carry information about the target or cause clutter for the received signal. O. Kilic EE 543

4. Boundary Conditions E, H n i Medium 1 t Medium 2 * Boundary conditions are a direct consequence of Maxwell’s equations. * They can be derived from the integral form by assuming an infinitesimal closed path, or an infinitesimal volume across the boundary. O. Kilic EE 543

5. Continuity of the Normal Component Medium 1 D1 n1 Medium 2 Dz 0 dA1 dA2 Medium 1 n2 Medium 2 D2 O. Kilic EE 543

6. Continuity of the Tangential Component Medium 1 t D1 Medium 2 Dz 0 t1 Medium 1 t2 Medium 2 D2 O. Kilic EE 543

7. Hn H Etan En n E General Form of Boundary Conditions E, H i n Medium 1 Medium 2 Components of E and H (Normal and Tangential): Htan n O. Kilic EE 543

8. Special Cases – Dielectric Boundary • For dielectric material or material with finite conductivity, Js = 0; i.e. surface current does not exist. Only volume current exists. Therefore: O. Kilic EE 543

9. Special Cases – Perfect Conductors • For perfect conductors, there can not be any voltage difference between any two points on the surface. Thus, Etan2 = 0 on the surface. • Also, due to the same reasoning there can not be any fields inside a perfect conductor. Therefore: O. Kilic EE 543

10. Example 1 A region contains a perfectly conducting half-space and air. We know that the surface current on the perfect conductor is . What is the tangentialH field in air just above the conductor? y H x Js O. Kilic EE 543

11. Plane of Incidence Defined by unit vectors i and n, where i is thedirection of the incident fields, and n is the unit normal vector to the boundary between the two medium. z In this example, x-z plane is the plane of incidence. n i x O. Kilic EE 543

12. Plane of Incidence and E, H Fields Incident fields Ei, Hi lie on a plane (P) perpendicular to i. Therefore, Ei, Hi can be decomposed into two basis (orthogonal) vectors that describe (P). There are infinitely many possible such pair of components. i H E P For instance, if i=z then x, y or x+y, x-y would be possible orthogonal vectors. O. Kilic EE 543

13. E and H Decomposition • We observe that E and H can be decomposed into infinitely many pairs of orthogonal vectors that lie on the plane P. • One choice is to have t = i x n as one component, where i: incidence direction, and n: normal to the boundary. • Since t is defined as the cross product of i with another vector, t is orthogonal to i, and thus can be a valid component of E or H. O. Kilic EE 543

14. Polarization • The selection of the orthogonal components for a reflection and transmission problem is usually done with respect to the plane of incidence. • These define two orthogonally polarized components of E and H fields. • These are called: • Perpendicular Polarization (TE, Horizontal) • Parallel Polarization (TM, Vertical) Defined for E field Defined for E field with respect to the interface Defined with respect to the plane of incidence O. Kilic EE 543

15. Perpendicular (TE) Polarization (Horizontal) • The electric field is perpendicular (transverse, TE) to the plane of incidence. Plane formed by i and n i n REMARK: Since n is on the plane of incidence, this condition (TE) implies that Elies on the interface; i.e. is horizontal. E O. Kilic EE 543

16. Parallel Polarization (TM) (Vertical) • The electric field does not have a component perpendicular to the plane of incidence; i.e. E lies on the plane of incidence. • Hence, the magnetic field is perpendicular (transverse, TM) to the plane of incidence. n REMARK: Since n is on the plane of incidence, this condition (TM) implies that Ehas a component perpendicular to the interface i.e. is vertical. i H O. Kilic EE 543

17. Decomposition into Two Polarizations Therefore, both E and H fields can be represented in the most general sense as a sum of these two orthogonal polarizations; i.e. TM (parallel) TE (perpendicular) O. Kilic EE 543

18. Example: Planar Boundary with a Dielectric Interface z E, H n: normal to the surface i: direction of propagation i n x O. Kilic EE 543

19. Reflection and Transmission – Perpendicular Polarization (TE) z Ei X o i n Hi qi qr Plane of incidence: x-z e1, m1 x e2, m2 qr t E is perpendicular to the plane of incidence. H lies on the plane of incidence. O. Kilic EE 543

20. Calculation of TE Coefficients • Due to the law of reflection (from matching the boundary conditions) • The incident, reflected and transmitted fields can be written in terms of the propagation direction, and reflection and transmission coefficients for TE waves. O. Kilic EE 543

21. TE Geometry Let the unit vectors along the direction of propagation be: z k1r i o incident q1 reflected q1 k1i transmitted Wave vectors: qt k2 t O. Kilic EE 543

22. Electric Field Expressions, TE Unknowns: O. Kilic EE 543

23. Magnetic Field Expressions, TE The corresponding magnetic fields are obtained from Maxwell’s equations where, O. Kilic EE 543

24. Application of Boundary Conditions Boundary conditions on the surface (dielectric medium): z Hr Ei o X i Hi Er X e1, m1 x e2, m2 Et X Ht t O. Kilic EE 543

25. BC for Electric Field O. Kilic EE 543

26. BC for Magnetic Field O. Kilic EE 543

27. Equations for TE In summary, the following has to be satisfied for all points across the boundary; i.e. for all x values. This implies that exponential terms have to be equal; i.e. Phase matching condition Snell’s Law O. Kilic EE 543

28. Solution for TE Thus, the boundary conditions result in: We obtain O. Kilic EE 543

29. Reflection and Transmission – Parallel Polarization (TM) Ei Hi X o i n qi qt e1, m1 x e2, m2 qr t H is perpendicular to the plane of incidence. E lies on the plane of incidence; i.e. is parallel to it. O. Kilic EE 543

30. Calculation of TM Coefficients • Due to duality principle, the reflection and transmission coefficients for TM can be obtained from the TE case by letting O. Kilic EE 543

31. Solution for TM HW Problem: Carry out the derivation for TM mode following similar steps as in TE case, and prove the equation above. O. Kilic EE 543

32. Power Coefficients • The R, T coefficients describe field (E,H) behavior. • The power reflection coefficient (reflectivity) G and transmission coefficient (transmissivity) T are defined with respect to the normal components of the time average Poynting vector. O. Kilic EE 543

33. Power Conservation Sav_z, inc Sav_z, ref Sav_x, inc Sav_x, inc Sav_x, ref Sav_x, ref Sav_x, tr Sav_x, tr Sav_z, tr Sav_z, ref Sav_z, inc Sav_z, tr = + O. Kilic EE 543

34. Reflectivity and Transmissivity where Time average Poynting Vector O. Kilic EE 543

35. Reflectivity and Transmissivity, TE Due to b.c. Note that the field coefficients satisfy: While the power coefficients satisfy energy conservation: Due to power conservation O. Kilic EE 543

36. Reflectivity and Transmissivity, TM Note that cos(q2) can be a complex number for certain incidence angles, such as in total reflection phenomenon. O. Kilic EE 543

37. Total Reflection (Critical Angle) • If the wave is incident from a dense to a less dense medium (i.e. k1>k2), it is possible to have no transmission into the second medium. • Recall Snell’s Law: >1 will result in pure imaginary q2 Can be > 1 for certain values of q1 O. Kilic EE 543

38. Transmission Angle for Incidence Beyond Critical Angle Pure imaginary O. Kilic EE 543

39. Critical Angle • The smallest incidence angle at which no real transmission angle exists is called the critical angle. • Thus all waves that arrive at incidence angles greater than the critical angle suffer total reflection. O. Kilic EE 543

40. Total Reflection for Incidence Beyond Critical Angle Therefore, for power considerations and O. Kilic EE 543

41. Example – Light source under water For an isotropic (i.e. equal radiation in all directions) light source, only light rays within a cone of qc can be transmitted to air. The permittivity of water at optical frequencies is 1.77eo. Calculate the value of the critical angle. k2 k1 O. Kilic EE 543

42. Total Transmission (Brewster Angle) • There exists an angle at which total transmission occurs for TM (parallel = vertically polarized) wave. O. Kilic EE 543

43. Proof O. Kilic EE 543

44. Total Transmission • Note that total transmission phenomenon only occurs for TM (parallel) polarization. • As a result, randomly polarized waves become polarized on reflection. (TM portion is not reflected back) • Note that: • for incidence angles smaller that qB, R// term is less than zero. • for incidence angles larger than qB, R// term is greater than zero. O. Kilic EE 543

45. Brewster Angle Impact on Polarization Consider circular polarization: Polarization changes direction when R// changes sign. Impact on Polarization O. Kilic EE 543

46. HW Problem • Plot the magnitude of reflection and transmission coefficients for a plane wave incident on a flat ground for incidence angles [0-90] degrees. Assume a lossless ground, i.e. zero conductivity and er2 = 4.0. Plot both TE and TM cases on the same chart. O. Kilic EE 543

47. qB Brewster vs Critical Angle Note that Brewster angle is always less than critical angle. qc O. Kilic EE 543

48. Example • A perpendicularly polarized em wave impinges from medium 1 to medium 2 as shown below. Calculate: x • the critical angle • kx, kz in terms of w, mo, eo • ktz in terms of w, mo, eo • Reflection and transmission coefficients e2, m2 e1, m1 z 60o Ei O. Kilic EE 543

49. About Lossy Media The Brewster angle loses its meaning if one of the media is lossy, in which case the Brewster angle θB will be complex-valued. O. Kilic EE 543

50. Solution (1/3) O. Kilic EE 543