# OpticsI09FresnelsEqn sNEW

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### Fresnel's Equations for Reflection and Refraction:

Fresnel's Equations for Reflection and Refraction Incident, transmitted, and reflected beams at interfaces Reflection and transmission coefficients The Fresnel Equations Brewster's Angle Total internal reflection Power reflectance and transmittance Phase shifts in reflection The mysterious evanescent wave

### Slide2:

Definitions: Planes of Incidence and the Interface and the polarizations Perpendicular (“S”) polarization sticks out of or into the plane of incidence. Plane of the interface (here the yz plane) (perpendicular to page) Plane of incidence (here the xy plane) is the plane that contains the incident and reflected k-vectors. ni nt qi qr qt Ei Er Et Interface Parallel (“P”) polarization lies parallel to the plane of incidence. Incident medium Transmitting medium

### Slide3:

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.

### Fresnel Equations:

Fresnel Equations We would like to compute the fraction of a light wave reflected and transmitted by a flat interface between two media with different refractive indices. where E0i, E0r, and E0t are the field complex amplitudes. We consider the boundary conditions at the interface for the electric and magnetic fields of the light waves. We’ll do the perpendicular polarization first. for the perpendicular polarization for the parallel polarization

### Boundary Condition for the Electric Field at an Interface:

The Tangential Electric Field is Continuous In other words: The total E-field in the plane of the interface is continuous. Here, all E-fields are in the z-direction, which is in the plane of the interface (xz), so: Ei(x, y = 0, z, t) + Er(x, y = 0, z, t) = Et(x, y = 0, z, t) ni nt qi qr qt Ei Bi Er Br Et Bt Interface Boundary Condition for the Electric Field at an Interface

### Boundary Condition for the Magnetic Field at an Interface:

The Tangential Magnetic Field* is Continuous In other words: The total B-field in the plane of the interface is continuous. Here, all B-fields are in the xy-plane, so we take the x-components: –Bi(x, y=0, z, t) cos(qi) + Br(x, y=0, z, t) cos(qr) = –Bt(x, y=0, z, t) cos(qt) *It's really the tangential B/m, but we're using m = m0 Boundary Condition for the Magnetic Field at an Interface ni nt qi qr qt Er Br Et Bt Interface qi qi

### Reflection and Transmission for Perpendicularly (S) Polarized Light:

Reflection and Transmission for Perpendicularly (S) Polarized Light Canceling the rapidly varying parts of the light wave and keeping only the complex amplitudes:

### Reflection & Transmission Coefficients for Perpendicularly Polarized Light:

Reflection & Transmission Coefficients for Perpendicularly Polarized Light

### Simpler expressions for r┴ and t┴:

Simpler expressions for r┴ and t┴ Recall the magnification at an interface, m: Also let r be the ratio of the refractive indices, nt / ni. Dividing numerator and denominator of r and t by ni cos(qi):

### Fresnel Equations—Parallel electric field:

Fresnel Equations—Parallel electric field Note that the reflected magnetic field must point into the screen to achieve . The x means “into the screen.” Note that Hecht uses a different notation for the reflected field, which is confusing! Ours is better! ni nt qi qr qt Ei Bi Er Br Et Bt Interface Beam geometry for light with its electric field parallel to the plane of incidence (i.e., in the page) This B-field points into the page.

### Reflection & Transmission Coefficients for Parallel (P) Polarized Light:

Reflection & Transmission Coefficients for Parallel (P) Polarized Light For parallel polarized light, B0i - B0r = B0t and E0icos(qi) + E0rcos(qr) = E0tcos(qt) Solving for E0r / E0i yields the reflection coefficient, r||: Analogously, the transmission coefficient, t|| = E0t / E0i, is These equations are called the Fresnel Equations for parallel polarized light.

### Simpler expressions for r║ and t║:

Simpler expressions for r║ and t║ Again, use the magnification, m, and the refractive-index ratio, r . And again dividing numerator and denominator of r and t by ni cos(qi):

### Reflection Coefficients for an Air-to-Glass Interface:

Reflection Coefficients for an Air-to-Glass Interface nair » 1 < nglass » 1.5 Note that: Total reflection at q = 90° for both polarizations Zero reflection for parallel polarization at Brewster's angle (56.3° for these values of ni and nt). (We’ll delay a derivation of a formula for Brewster’s angle until we do dipole emission and polarization.)

### Reflection Coefficients for a Glass-to-Air Interface:

Reflection Coefficients for a Glass-to-Air Interface nglass » 1.5 > nair » 1 Note that: Total internal reflection above the critical angle qcrit º arcsin(nt /ni) (The sine in Snell's Law can't be > 1!): sin(qcrit) = nt /ni sin(90)

### Transmittance (T):

Transmittance (T) T º Transmitted Power / Incident Power A = Area Compute the ratio of the beam areas: The beam expands in one dimension on refraction. The Transmittance is also called the Transmissivity. 1D beam expansion

### Reflectance (R):

Reflectance (R) R º Reflected Power / Incident Power 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: A = Area The Reflectance is also called the Reflectivity.

### Reflectance and Transmittance for an Air-to-Glass Interface:

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

### Reflectance and Transmittance for a Glass-to-Air Interface:

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

### Reflection at normal incidence:

Reflection at normal incidence When qi = 0,   and   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 mirrors 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: Practical Applications of Fresnel’s Equations Optical fibers use total internal reflection. Hollow fibers use high-incidence-angle near-unity reflections.

### Phase Shift in Reflection (for Perpendicularly Polarized Light):

Phase Shift in Reflection (for Perpendicularly Polarized Light) 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):

Phase Shift in Reflection (Parallel Polarized Light) 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 qi = 0! Also, the phase is opposite above Brewster’s angle.

### Phase shifts in reflection (air to glass):

Phase shifts in reflection (air to glass) 180° phase shift for all angles 180° phase shift for angles below Brewster's angle; 0° for larger angles

### Phase shifts in reflection (glass to air):

Phase shifts in reflection (glass to air) Interesting phase above the critical angle 180° phase shift for angles below Brewster's angle; 0° for larger angles

### Phase shifts vs. incidence angle and ni /nt:

Phase shifts vs. incidence angle and ni /nt Li Li, OPN, vol. 14, #9, pp. 24-30, Sept. 2003 ni /nt ni /nt Note the general behavior above and below the various interesting angles… qi qi

### If you slowly turn up a laser intensity incident on a piece of glass, where does damage happen first, the front or the back?:

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. But constructive interference happens at the back surface between the incident light and the reflected wave. This yields an irradiance that is 44% higher just inside the back surface!

### Phase shifts with coated optics:

Phase shifts with coated optics 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. Example: Laser Mirror

### Total Internal Reflection occurs when sin(qt) > 1, and no transmitted beam can occur.:

Total Internal Reflection occurs when sin(qt) > 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. Total internal reflection is 100% efficient, that is, all the light is reflected. Brewster’s angle Total Internal Reflection

### Applications of Total Internal Reflection:

Applications of Total Internal Reflection Beam steerers Beam steerers used to compress the path inside binoculars

### Three bounces: The Corner Cube:

Three bounces: The Corner Cube Corner cubes involve three reflections and also displace the return beam in space. Even better, they always yield a parallel return beam: Hollow corner cubes avoid propagation through glass and don’t use TIR. If the beam propagates in the z direction, it emerges in the –z direction, with each point in the beam (x,y) reflected to the (-x,-y) position.

### Fiber Optics:

Optical fibers use TIR to transmit light long distances. Fiber Optics 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 Design of optical fibers ncore > ncladding

### Propagation of light in an optical fiber:

Propagation of light in an optical fiber 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. 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.

### Microstructure fiber:

Microstructure fiber Such fiber has many applications, from medical imaging to optical clocks. Photographs courtesy of Jinendra Ranka, Lucent Air holes Core In microstructure fiber, air holes act as the cladding surrounding a glass core. Such fibers have different dispersion properties.

### Frustrated Total Internal Reflection:

Frustrated Total Internal Reflection By placing another surface in contact with a totally internally reflecting one, total internal reflection can be frustrated. 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. n n n n Total internal reflection Frustrated total internal reflection n=1 n=1

### FTIR and fingerprinting:

FTIR and fingerprinting See TIR from a fingerprint valley and FTIR from a ridge.

### The Evanescent Wave:

The Evanescent Wave The evanescent wave is the "transmitted wave" when total internal reflection occurs. A mystical quantity! So we'll do a mystical derivation:

### The Evanescent-Wave k-vector:

The Evanescent-Wave k-vector The evanescent wave k-vector must have x and y components: Along surface: ktx = kt sin(qt) Perpendicular to it: kty = kt cos(qt) Using Snell's Law, sin(qt) = (ni /nt) sin(qi), so ktx is meaningful. And again: cos(qt) = [1 – sin2(qt)]1/2 = [1 – (ni /nt)2 sin2(qi)]1/2 = ± ib Neglecting the unphysical -ib solution, we have: Et(x,y,t) = E0 exp[–kb y] exp i [ k (ni /nt) sin(qi) x – w t ] The evanescent wave decays exponentially in the transverse direction.

### Optical Properties of Metals:

Optical Properties of Metals A simple model of a metal is a gas of free electrons (the Drude model). These free electrons and their accompanying positive nuclei can undergo "plasma oscillations" at frequency, wp. where:

### Reflection from metals:

Reflection from metals At normal incidence in air: Generalizing to complex refractive indices: