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Home , Fresnel equations, Plane of incidence

Fresnel's Equations for and

Incident, transmitted, and reflected beams at interfaces

Reflection and transmission coefficients

The

Brewster's Angle

Total internal reflection

Power and transmittance

Phase shifts in reflection

The mysterious evanescent wave Reflection and transmission for an arbitrary angle of incidence at one (1) interface

• Only Maxwell+Boundary conditions need. Gives Fresnels equations Maxwell’s eqns.

r D =⋅∇ ρext r B =⋅∇ 0 r r ∂B D −=×∇ dt r r ∂D r H =×∇ + j dt Boundary conditions of EM wave

• Tangential components of the: - E and H fields (from Gauss’ theorem) • components of - D and B fields (from Stoke’s theorem)

r r r )1()2( 12 EEn =−× 0)( r rr )1()2( 1 r 2 12 BBn =−⋅ 0)( n12 Definitions: Planes of Incidence and the Interface and the polarizations

Perpendicular (“S”) Incident medium sticks out of or into the plane of r r incidence. ki kr E Ei r ni (here the xy plane) is the plane that θi θr contains the incident InterfacePlane of the interface (here the and reflected k- yz plane)y (perpendicular to page) vectors. θt x nt z E r Parallel (“P”) t k polarization lies t parallel to the plane of incidence. Transmitting medium Shorthand notation for the polarizations

Perpendicular (S) polarization sticks up out of the plane of incidence.

Parallel (P) polarization lies parallel to the plane of incidence. rEErEE== // Fresnel Equations For example, ⊥ 0000riri tEEtEE== // ⊥ 0000titi We would like to compute the fraction of a wave reflected and transmitted by a flat interface between two media with different refrac- tive indices. Fresnel was the first to do this calculation.

It proceeds by considering r r the boundary ki kr conditions at Er Ei ni the interface Bi θ θ Br for the electric i r y Interface and magnetic Beam geometry fields of the for light with its z x θ light waves. per- t E pendicular to the t nt We’ll do the plane of incidence r (i.e., out of the page) Bt perpendicular kt polarization first. Boundary Condition for the Electric y Field at an Interface

The Tangential Electric Field is Continuous z x

In other words: r r The total E-field in ki kr Er the plane of the Ei ni interface is B B i θi θr r continuous. Interface Here, all E-fields are θt in the z-direction, E t nt which is in the plane B r of the interface (xz), t kt so:

Ei(x, y = 0, z, t) + Er(x, y = 0, z, t) = Et(x, y = 0, z, t) Boundary Condition for the Magnetic y Field at an Interface z x The Tangential Magnetic Field* is Continuous

In other words: r r k The total B-field in i kr Er the plane of the Ei ni interface is θi B B θ θi θr r continuous. i i Interface Here, all B-fields are θ in the xy-plane, so t Et n we take the t B r x-components: t kt

–Bi(x, y=0, z, t) cos(θi) + Br(x, y=0, z, t) cos(θr) = –Bt(x, y=0, z, t) cos(θt)

*It's really the tangential B/μ, but we're using μ = μ0 Reflection and Transmission for Perpendicularly (S) Polarized Light

Canceling the rapidly varying parts of the light wave and keeping only the complex amplitudes:

EE00ir+ = E 0 t

−+BB00iirr cos(θ ) cos(θθ ) =− B 0 tt cos( )

But B= E/( c00 / n )== nE / c and θriθ :

nEir()cos()cos()00− E iθ i=− nE tt0 θ t

Substituting forE0000tirt using EE+ = E :

nEiriitrit()cos()()cos()00− Eθ =− nE00 + E θ Reflection & Transmission Coefficients for Perpendicularly Polarized Light

Rearranging nEiriitrit()cos()()cos()00− Eθ =− nE00 + E θ yields:

En00ri[]cos(θ i )+= n t cos(θθθ t ) En ii[] cos( i ) − n t cos( t )

Solving for EE00ri/ yields the :

rEE⊥ ==00riiittiitt/[ n cos(θ ) − n cos(θθθ )] /[ n cos( )+ n cos( )]

Analogously, the transmission coefficient, EE00ti/ , is

tEE⊥ ==00tiiiiitt/ 2 n cos(θ ) /[ n cos(θθ )+ n cos( )]

These equations are called the Fresnel Equations for perpendicularly polarized light. Simpler expressions for r┴ and t┴

Recall the magnification at

an interface, m: θi wi ni w cos(θ ) nt m ==tt w θ t wiicos(θ ) t

Also let ρ be the ratio of the refractive indices, nt / ni.

Dividing numerator and denominator of r and t by ni cos(θi): rn⊥ =−[ iittiittcos(θ ) n cos(θθθρρ )] /[ n cos( ) +=−+ n cos( )] [ 1 m] /[ 1 m]

tn⊥ =+=2iiiitt cos(θ ) /[ n cos(θθ ) n cos( )] 2 /[ 1 ρ+m]

1− ρm 2 r = t = ⊥ 1+ ρm ⊥ 1+ ρm Fresnel Equations—Parallel electric field

r r This B-field y ki kr points into E B the page. i r z x Bi E θi θr r ni Interface

Beam geometry Note that Hecht for light with its θ uses a different electric field t E notation for the parallel to the t nt reflected field, plane of incidence B which is t r confusing! (i.e., in the page) kt Ours is better!

Note that the reflected magnetic field must point into the screen to rr r achieve EB ×∝ k . The x means “into the screen.” Reflection & Transmission Coefficients for Parallel (P) Polarized Light

For parallel polarized light, B0i -B0r = B0t

and E0icos(θi) + E0rcos(θr) = E0tcos(θt)

Solving for E0r / E0i yields the reflection coefficient, r||:

rEE||== 0riittiitti/ 0 [ n cos(θ ) − n cos(θθθ )] /[ n cos( )+ n cos( )]

Analogously, the transmission coefficient, t|| = E0t / E0i, is

tEE||== 0tiiiitti/ 0 2 n cos(θ ) /[ n cos(θθ )+ n cos( )]

These equations are called the Fresnel Equations for parallel polarized light. Simpler expressions for r║ and t║

rEE||== 0riittiitti/ 0 [ n cos(θ ) − n cos(θθθ )] /[ n cos( )+ n cos( )]

tEE||== 0ti/ 0 2 n i cos(θ i ) /[ n i cos(θθ t )+ n t cos( i )]

Again, use the magnification, m, and the refractive-index ratio, ρ .

r And again dividing numerator and denominator of r and t by ni cos(θi):

rm= −+ρ / mρ r|| [ ] [ ]

ttm= 2/ + ρ || [ ]

m − ρ 2 rr||= tt = m + ρ || m + ρ Reflection Coefficients for an Air-to-Glass Interface

1.0 nair ≈ 1 < nglass ≈ 1.5

Note that: Brewster’s angle .5 r =0! Total reflection at θ = 90° || r|| for both polarizations

0 Zero reflection for parallel polarization at Brewster's angle (56.3° for these Reflection coefficient, r -.5 r values of ni and nt). ┴

-1.0 0° 30° 60° 90°

Incidence angle, θi Condition for rII (rp)=0

θ1 + θ2 = 90°

⎛ n ⎞ −1⎜ t ⎟ θ B = tan ⎜ ⎟ ⎝ ni ⎠

Brewster’s angle

[pic from Wikipedia] Brewster application: Sunglasses

• Polarized sunglasses use the principle of Brewster's angle to eliminate glare from from the sun. In a large range of angles around Brewster's angle the reflection of p-polarized light is lower than s-polarized light.

• Thus, if the sun is low in the sky mostly s-polarized light will reflect from water. Sunglasses made up of (e.g. polaroid film) aligned to block this light consequently block reflections from the water. To accomplish this, sunglass makers assume people will be upright while viewing the water and thus align the polarizers to block the polarization which oscillates along the line connecting the sunglass ear-pieces (i.e. Horizontal).

• Photographers use the same principle to remove reflections from water so that they might photograph objects beneath the surface. In this case, the filter camera attachment can be rotated to be at the correct angle (see figure). Brewster angle

Polarizer aligned in ”s” direction Polarizer aligned in ”p” direction

Photographs taken of mudflats with a camera polarizer filter rotated to two different angles. In the first picture, the polarizer is rotated to maximize reflections, and in the second, it is rotated 90° to minimize reflections - almost all reflected sunlight is eliminated [pic from Wikipedia] Reflection Coefficients for a Glass-to-Air Interface

1.0 nglass ≈ 1.5 > nair ≈ 1 Critical angle .5 r Note that: ┴ Total internal reflection Total internal reflection above the critical angle 0 Brewster’s -1 angle θcrit ≡ sin (nt /ni) Reflection coefficient, r -.5 (The sine in Snell's Law Critical r|| can't be > 1!): angle -1.0 0° 30° 60° 90° sin(θcrit) = nt /ni sin(90°) Incidence angle, θi Relation between θB and θC

• Note tanθB=sinθC, tanθ>sinθ⇒θC>θB

θB θc θi ⎛⎞ε 00c 2 Transmittance (T) In= ⎜⎟ E0 ⎝⎠2 IA T ≡ Transmitted Power / Incident Power = tt A = Area IAii

θ Compute the w i i n Awcos(θ ) ratio of the i tt= == tm nt beam areas: Awiicos(θ i ) 1D beam wt expansion θt The beam expands in one dimension on refraction. 2 E0t 2 2 = t ε c 2 ⎛⎞00 E0i ⎜⎟nEtt0 2 IAtt ⎝⎠2 ⎡⎤wt nEtt0 w tnt2 cos(θ t ) Tt== ⎢⎥=2 = IA ε c 2 w n cos(θ ) ii ⎛⎞00 ⎣⎦i nEii0 w i i i ⎜⎟nEii0 ⎝⎠2

⎡⎤(nttcos(θ )) The Transmittance is also ⇒ Tt==⎢⎥22ρ mt n cos θ called the Transmissivity. ⎣⎦⎢⎥()ii() Reflectance (R) ⎛⎞ε 00c 2 In= ⎜⎟ E0 ⎝⎠2 IA R ≡ Reflected Power / Incident Power = rr A = Area IAii

θ θ w i r i wi ni nt

Because the angle of incidence = the angle of reflection, the beam area doesn’t change on reflection.

Also, n is the same for both incident and reflected beams.

So: R = r 2 The Reflectance is also called the Reflectivity. Reflectance and Transmittance for an Air-to-Glass Interface

Perpendicular polarization Parallel polarization

1.0 1.0 T T

.5 .5

R R 0 0 0° 30° 60° 90° 0° 30° 60° 90°

Incidence angle, θi Incidence angle, θi

Note that R + T = 1 Reflectance and Transmittance for a Glass-to-Air Interface

Perpendicular polarization Parallel polarization

1.0 1.0 R R

.5 .5 T T 0 0 0° 30° 60° 90° 0° 30° 60° 90°

Incidence angle, θi Incidence angle, θi

Note that R + T = 1 Reflection at normal incidence

When θi = 0, 2 ⎛⎞nn− R = ⎜⎟ti ⎝⎠nnti+

and 4nnti T = 2 ()nnti+

For an air-glass interface (ni = 1 and nt = 1.5),

R = 4% and T = 96%

The values are the same, whichever direction the light travels, from air to glass or from glass to air.

The 4% has big implications for photography lenses. Practical Applications of Fresnel’s Equations

Windows look like at night (when you’re in the brightly lit room)

One-way mirrors (used by police to interrogate bad guys) are just partial reflectors (actually, aluminum-coated).

Disneyland puts ghouls next to you in the haunted house using partial reflectors (also aluminum-coated).

Lasers use Brewster’s angle components to avoid reflective losses:

R = 100% 0% reflection! medium R = 90%

0% reflection!

Optical fibers use total internal reflection. Hollow fibers use high- incidence-angle near-unity reflections. Shift in Reflection (for Perpendicularly Polarized Light) r r rEE⊥ = 00ri/ = ki kr nncos(θ )− cos(θ ) E []iitt Ei r n B θ θ i []nniittcos(θ )+ cos(θ ) i i r Br Interface [nn− ] When θ ==0, r it θ i t E []nnit+ t n r t Bt kt If nnit< () air to glass , r< 0

So there will be destructive interference between the incident and reflected beams just before the surface.

Analogously, if ni > nt (glass to air), r⊥ > 0, and there will be constructive interference. Phase Shift in Reflection (Parallel Polarized Light) r r k k rEE= 00ri/ = i r E Br i ni []nnitticos(θθ )− cos( ) B i θiθr Er []nnitticos(θθ )+ cos( ) Interface []nn− When θ ==0, r it θ i t E []nnit+ t nt Bt r If nnit< (air to glass) , r< 0 kt

This also means destructive interference with incident beam.

Analogously, if ni > nt (glass to air), r|| > 0, and we have constructive interference just above the interface.

Good that we get the same result as for r⊥; it’s the same problem when θi = 0! Also, the phase is opposite above Brewster’s angle. Phase shifts in reflection (air to glass)

ni < nt π

180° phase shift ┴ for all angles

0 0° 30° 60° 90° Incidence angle π 180° phase shift for angles below || Brewster's angle; 0° for larger angles 0 0° 30° 60° 90° Incidence angle Phase shifts in reflection (glass to air)

nt < ni π Interesting phase above the critical ┴ angle

0 0° 30° 60° 90° Incidence angle π 180° phase shift for angles below || Brewster's angle; 0° for larger angles 0 0° 30° 60° 90° Incidence angle Phase shifts vs. incidence angle and ni /nt

θi Note the general behavior ni /nt above and below the various interesting angles…

ni /nt Li Li, OPN, vol. 14, #9, pp. 24-30, Sept. 2003 θi If you slowly turn up a laser intensity incident on a piece of glass, where does damage happen first, the front or the back? The obvious answer is the front of the object, which sees the higher intensity first.

r

But constructive interference happens at the back surface between the incident light and the reflected wave.

This yields an that is 44% higher just inside the back surface! 2 (1+= 0.2) 1.44 (glass to air), r|| > 0 ⇒ r|| =[(1.5-1)/(1.5+1)] Phase shifts with coated

Reflections with different magnitudes can be generated using partial metallization or coatings. We’ll see these later.

But the phase shifts on reflection are the same! For near-normal incidence: 180° if low-index-to-high and 0 if high-index-to-low.

Highly reflecting coating Example: on this surface Laser

Phase shift of 180° Total Internal Reflection occurs when sin(θt) > 1, and no transmitted beam can occur.

Note that the irradiance of the transmitted beam goes to zero (i.e., TIR occurs) as it grazes the surface.

Brewster’s angle

Total Internal Reflection

Total internal reflection is 100% efficient, that is, all the light is reflected. Applications of Total Internal Reflection

Beam steerers

Beam steerers used to compress the path inside binoculars Fiber Optics

Optical fibers use TIR to transmit light long distances.

They play an ever-increasing role in our lives! Design of optical fibers

Core: Thin glass center of the fiber that carries the light

Cladding: Surrounds the core and reflects the light back into the core

Buffer coating: Plastic protective coating

ncore > ncladding Propagation of light in an optical fiber

Light travels through the core bouncing from the reflective walls. The walls absorb very little light from the core allowing the light wave to travel large distances.

Some signal degradation occurs due to imperfectly constructed glass used in the cable. The best optical fibers show very little light loss -- less than 10%/km at 1,550 nm.

Maximum light loss occurs at the points of maximum curvature. Microstructure fiber

Air holes In microstructure fiber, air holes act as the cladding surrounding a glass core. Such fibers have different dispersion properties.

Core

Such fiber has many applications, from medical imaging to optical clocks.

Photographs courtesy of Jinendra Ranka, Lucent Frustrated Total Internal Reflection

By placing another surface in contact with a totally internally reflecting one, total internal reflection can be frustrated.

Total internal reflection Frustrated total internal reflection n=1 n=1 n n n n

How close do the prisms have to be before TIR is frustrated?

This effect provides evidence for evanescent fields—fields that leak through the TIR surface–and is the basis for a variety of spectroscopic techniques. FTIR and fingerprinting

See TIR from a fingerprint valley and FTIR from a ridge. The Evanescent Wave r θ r k i k ni The evanescent wave is the i r y "transmitted wave" when total internal θt r k nt reflection occurs. A mystical quantity! x t So we'll do a mystical derivation:

E0r [nniittcos(θ )− cos(θ )] r⊥ == En0ii[]cos(θ itt )+ n cos(θ )

Since sin(θtt)1,> θ doesn't exist, so computing r⊥ is impossible.

Let's check the reflectivity, R , anyway. Use Snell's Law to eliminate θt :

2 22⎛⎞ni cos(θθtt )=− 1 sin ( ) =− 1⎜⎟ θ sin ( i ) =Neg. Number ⎝⎠nt

Substituting this expression into the above one for r⊥ and

* ⎛⎞⎛⎞abiabi−+ redefining RR yields: ≡=r⊥⊥r⎜⎟⎜⎟=1 ⎝⎠⎝⎠abiabi+− So all power is reflected; the evanescent wave contains no power. The Evanescent-Wave k-vector

The evanescent wave k-vector must have x and y components: r r θi ni ki kr

Along surface: ktx = kt sin(θt) y θt r k nt x t Perpendicular to it: kty = kt cos(θt)

Using Snell's Law, sin(θt) = (ni /nt) sin(θi), so ktx is meaningful.

2 1/2 2 2 1/2 And again: cos(θt) = [1 – sin (θt)] = [1 – (ni /nt) sin (θi)]

= ±iβ

Neglecting the unphysical -iβ solution, we have:

Et(x,y,t) = E0 exp[–kβ y] exp i [ k (ni /nt) sin(θi) x – ω t ]

The evanescent wave decays exponentially in the transverse direction. Optical Properties of Metals

A simple model of a metal is a gas of free electrons

These free electrons and their accompanying positive nuclei can

undergo "plasma oscillations" at frequency, ωp.

2 where: 2 Ne ωp = ()ε0me

The for a metal is: 2 2 ⎛⎞ωp n =− 1 ⎜⎟ ⎝⎠ω

When nn2 < 0, is imaginary, and absorption is strong.

So for ωω<>pp metals absorb strongly. For ωω metals are transparent. Reflection from metals (1)n − 2 At normal incidence in air: R = (1)n + 2 (1)(1)nn− * − Generalizing to complex refractive indices: R = (1)(1)nn+ * +