02941 Physically Based Rendering Reflection and Transmission Jeppe - PowerPoint PPT Presentation

02941 Physically Based Rendering Reflection and Transmission Jeppe Revall Frisvad June 2016

Multiscale light modelling ◮ Light at different scales: ◮ Quantum electrodynamics (photons) ◮ Electromagnetic radiation (waves) ◮ Geometrical optics (rays) ◮ Radiative transfer (ray bundles) ◮ Why do we need to know? To understand: ◮ how light interacts with materials; ◮ where the material properties come from; ◮ when rays don’t work. ◮ Geometrical optics include: ◮ the law of reflection (Euclid ∼ 300 B.C. and before); ◮ the law of refraction (Ibn Sahl ∼ 984 , Snel van Royen 1621 ). ◮ Electromagnetic radiation (Maxwell 1873 ) includes: ◮ the index of refraction; ◮ the amount of reflection and transmission at a specular surface (Fresnel 1832 ).

Time-harmonic Maxwell equations for isotropic materials ∇ × H c ( σ − i ωε ) E c = ∇ × E c = i ωµ H c ∇ · ( ε E c ) = 0 ∇ · ( µ H c ) = 0 , where ◮ ω = 2 π c /λ is the angular frequency of the light ◮ Re ( E c ) is the electric field vector ◮ Re ( H c ) is the magnetic vector ◮ and the following are the isotropic material properties : ◮ σ is the conductivity ◮ ε is the permittivity ◮ µ is the permeability

The plane wave solution ◮ Reflection and refraction occurs at a surface marking the boundary between two (at least locally) homogeneous materials. ◮ Plane waves satisfy the time-harmonic Maxwell equations for isotropic and homogeneous materials. ◮ The plane wave equations are: E 0 e − i ( ω t − k · x ) E c ( x , t ) = H 0 e − i ( ω t − k · x ) . H c ( x , t ) = where ◮ x is a position in space ◮ t is time ◮ E 0 and H 0 are wave amplitudes ◮ k is the wave vector.

Plane wave Maxwell equations ◮ The plane wave solution inserted in the time-harmonic Maxwell equations for isotropic, homogeneous materials: k × H 0 = − ω ( ε + i σ/ω ) E 0 k × E 0 = ωµ H 0 k · E 0 = 0 k · H 0 = 0 . ◮ This boils down to the following conditions that plane waves must satisfy to be physically realisable: k · E 0 = k · H 0 = E 0 · H 0 = 0 k · k = ω 2 µ ( ε + i σ/ω ) . ◮ Thus k is a complex vector if the material is a conductor.

The meaning of the wave vector ◮ Consider the exponential part of the plane wave equation: e − i ( ω t − k · x ) = e − i ω t e i k · x ◮ The first part concerns temporal aspects, the second part concerns spatial aspects. ◮ The wave vector k determines the spatial aspect (the wave propagation). ◮ If the material is a conductor, k = k ′ + i k ′′ is complex: e i k · x = e i k ′ · x e − k ′′ · x . ◮ This reveals that the phase velocity is v = ω/ k ′ and ◮ k ′ is normal to the surface of constant phase ◮ k ′′ is normal to the surface of constant amplitude ◮ They determine the direction and damping of the wave.

The index of refraction (IOR) ◮ Recall the only remaining condition which involves material properties: k · k = ω 2 µ ( ε + i σ/ω ) . ◮ This is a golden opportunity to reduce the number of material properties. ◮ Suppose we introduce n = n ′ + in ′′ = c � µ ( ε + i σ/ω ) . ◮ Then the condition becomes k · k = ω 2 c 2 n 2 , where ◮ c is the speed of light in vacuo . ◮ n is called the (complex) index of refraction.

Phase velocity and amplitude damping ◮ A closer look at the condition which describes the relation between material and wave propagation: k · k = k ′ · k ′ − k ′′ · k ′′ + i 2 k ′ · k ′′ = ω 2 c 2 n 2 . ◮ For materials that are not strong absorbers: k ′′ · k ′′ ≈ 0. ◮ Then (equating the real parts): k ′ ≈ ω v = ω k ′ ≈ c c n ′ ⇒ n ′ the phase velocity v is determined by the real part of the IOR. ◮ And (equating the imaginary parts): 2 k ′ k ′′ cos θ = ω 2 n ′′ k ′′ ≈ ω cos θ ≈ ω c n ′′ , c 2 2 n ′ n ′′ ⇒ c where θ (usually close to 0) is the angle between k ′ and k ′′ .

The Poynting vector and absorption ◮ The Poynting vector S determines the direction and magnitude of energy propagation in an electromagnetic field. ◮ For the plane wave solution, we have: S = ε 0 c 2 µ Re ( E c ) × Re ( H c ) . ◮ Inserting the plane wave equations and taking the magnitude of S determines the absorption of energy by the material: � � � e − 2 k ′′ · x . | S | = ε 0 c 2 µ � Re ( E 0 e − i ( ω t − k ′ · x ) ) × Re ( H 0 e − i ( ω t − k ′ · x ) ) � � ◮ This reveals that energy is exponentially attenuated by 2 k ′′ , or c n ′′ = 4 π n ′′ σ a = 2 k ′′ ≈ 2 ω , λ where σ a is called the absorption coefficient and λ is the wavelength in vacuo .

Plane waves incident on surfaces ◮ Consider a plane wave incident on a smooth surface. ◮ The wave gives rise to two other waves. Altogether we have ◮ An incident wave (subscript i ) ◮ A reflected wave (subscript r ) ◮ A transmitted wave (subscript t ) ◮ We resolve all waves into two independent components: ◮ The wave with the electric vector perpendicular to the plane of incidence: ⊥ -polarised light. ◮ The wave with the electric vector parallel to the plane of incidence: � -polarised light. ◮ Maxwell’s equations require that the field vectors are continuous across the surface boundary. Let us call this the continuity condition . ◮ The continuity condition must hold at all times and no matter where we place the point of incidence in space.

Conditions imposed by Maxwell’s equations ◮ Using the continuity condition: E ⊥ i + E ⊥ r = E ⊥ t . ◮ This must also hold for x = 0, therefore 0 i e − i ω i t + E ⊥ 0 r e − i ω r t = E ⊥ 0 t e − i ω t t . E ⊥ ◮ This is true only if ω i = ω r = ω t . ◮ With the condition k · k = n 2 ω 2 / c 2 , this reveals k i · k i = k r · k r = k t · k t n 2 n 2 n 2 t i i which shows how the index of refraction governs the propagation of plane waves of light.

The plane of incidence z Ei Er k x,i k x,r k z,i k z,r Θ i Θ r ni x nt k x,t Θ t k z,t Et ◮ At the boundary, z = 0, we have at the time t = 0: 0 i e i ( k x , i x + k y , i y ) + E ⊥ 0 r e i ( k x , r x + k y , r y ) = E ⊥ E ⊥ 0 t e i ( k x , t x + k y , t y ) which must hold for all x and y .

The plane of incidence z Ei Er k x,i k x,r k z,i k z,r Θ i Θ r ni x nt k x,t Θ t k z,t Et ◮ The result is k x , i = k x , r = k x , t k y , i = k y , r = k y , t . and

The plane of incidence ◮ The result is k x , i = k x , r = k x , t and k y , i = k y , r = k y , t . ◮ Maxwell’s equations require that k is perpendicular to E 0 . ◮ Since E ⊥ 0 i (by definition) is perpendicular to the plane of incidence, k i must be parallel to the plane of incidence. ◮ Then k y , i = 0. ◮ Meaning that k y , r = k y , t = k y , i = 0. ◮ In other words, the reflected and transmitted vectors lie in the plane of incidence .

The law of reflection ◮ Using the following results (found previously): k i · k i = k r · k r n 2 n 2 i i k x , i = k x , r and k y , r = k y , i = 0 . ◮ We find: k 2 z , i = k 2 z , r with the solutions k z , r = k z , i or k z , r = − k z , i . ◮ Only the solution k z , r = − k z , i describes a wave on the right side of the surface. ◮ The law of reflection is then: The reflected wave lies in the plane of incidence, the (complex) angle of reflection is equal to the (complex) angle of incidence .

The law of refraction ◮ Using the following results (found previously): k i · k i = k t · k t n 2 n 2 t i k x , i = k x , t ◮ We find: k x , i k x , t n i √ k i · k i = n t √ k t · k t or (using complex angles) n i sin Θ i = n t sin Θ t . ◮ This is called generalised Snell’s law . ◮ The law of refraction is then: The refracted wave lies in the plane of incidence, the (complex) angle of refraction follows the generalised Snell’s law .

Continuity of the magnetic vector ◮ One of the plane wave Maxwell equations is: H 0 = 1 ωµ k × E 0 . ◮ The x component of this equation is H 0 x = 1 ωµ ( k y E 0 z − k z E 0 y ) . ◮ Using the perpendicular components and the result that k y , i = k y , r = k y , t = 0, we have H 0 x = − k z (0 , E 0 y , 0) = E ⊥ ωµ E 0 y and . 0 ◮ The continuity condition (invoked on the x component of the magnetic vector) is then − k z , i 0 i − k z , r 0 r = − k z , t E ⊥ E ⊥ E ⊥ 0 t . ω i µ i ω r µ i ω t µ t

The Fresnel equations for reflection ◮ The newly found continuity condition: − k z , i 0 i − k z , r 0 r = − k z , t E ⊥ E ⊥ E ⊥ 0 t . ω i µ i ω r µ i ω t µ t ◮ Using the law of reflection k z , r = − k z , i and the result that angular frequencies are equal ω i = ω r = ω t , we have µ i k z , i E ⊥ 0 i − k z , i E ⊥ E ⊥ 0 r = k z , t 0 t . µ t ◮ Ignoring magnetic effects, we set µ i /µ t = 1. ◮ Using the other continuity condition, E ⊥ 0 i + E ⊥ 0 r = E ⊥ 0 t , we get 0 r = k z , i − k z , t k z , i E ⊥ 0 i − k z , i E ⊥ 0 r = k z , t ( E ⊥ 0 i + E ⊥ E ⊥ E ⊥ 0 r ) ⇔ 0 i . k z , i + k z , t

The Fresnel equations for reflection ◮ The result we arrived at was 0 r = k z , i − k z , t E ⊥ E ⊥ 0 i . k z , i + k z , t √ k i · k i and using cos Θ = k z / √ k i · k i , we find the amplitude ratio ◮ Diving by E ⊥ 0 i r ⊥ = E ⊥ = n i cos Θ i − n t cos Θ t 0 r ˜ . E ⊥ n i cos Θ i + n t cos Θ t 0 i ◮ In a similar way, we can find r � = E � = n t cos Θ i − n i cos Θ t 0 r ˜ . E � n t cos Θ i + n i cos Θ t 0 i ◮ These are called Fresnel’s equations for reflection.