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RAY THEORY OF TRANSMISSION: Acceptance AngleSubmitted by:Akhil AroraM.E. [E.C.E.]Modular 2014-17IInd Spell Roll No. 141602Acceptance ConeAcceptance Angle, NATURE OF LIGHTConcept Light can be explained asRays: using OpticalGeometryWaves: using Electromagnetic TheoryPhotons: using PhotoelectricEffectWe will needOptical Geometry to explain light propagationElectromagnetic Theory to understand spectrumPhotoelectric Effect to show lasers and photodetectors

NATURE OF LIGHTSpectrum Light as a wave

NATURE OF LIGHTPolarization Light as a waveAn electromagnetic wave has electric (E) and magnetic (H) fields. They are perpendicular

The orientation and phase of E defines the polarization (linear, circular or elliptical)

Circular polarizationPHYSICS OF LIGHTThe propagation of light can be analyzed in detail using electromagnetic wave theory. Light falls in the general category of electromagnetic waves, much like radio waves. The behavior of light is sometimes easier to explain by using ray tracings The propagation of light in a fiber can be described in terms of rays.

REFLECTIONREFLECTION :When a light ray is incident on a reflecting surface, the ray bounces back like a handball when it hits a wall.A reflecting surface is one that is highly polished, opaque and coated with special reflective materials.The law of reflection states that the angle of incidence is equal to the angle of reflection. The incident ray is the line AO, the reflected ray is OB and ON is the normal to the reflecting surface. The incident and reflected angles, 1 and 2, respectively, are those between the rays and the line perpendicular to the surface.ReflectionLED or LASER SourceREFRACTIONWhen a ray travels across a boundary between two materials with different refractive indices n1 and n2, both refraction and reflection takes place. The case where n1 > n2; that is where the light travels from high to low refractive index materials. The refracted ray is broken that is, the angle 2 is not equal to 1. The relation between 1 and 2 is given by Snells law of refraction.

RefractionLED or LASER SourceLIGHT REFRACTEDLower Refractive index RegionThis end travels further than the other handHigher Refractive index RegionREFRACTIVE INDEXRefractive Index of a mediumMeasures how much the speed of light is reduced

n typical value: 1.5

i: materials relative permittivityREAL in fiber (dielectric) i = j ; negligible in a dielectric

c 3108 m/s, light speed in vacuumMaximum speed of any physical phenomenac = 299,792,458 m/s (set by definition)

REFRACTIVE INDEXSnells Laws (Optical Geometry)Light = raysLight propagation straight-line (homogeneous medium)

(In refractionthere is alwaysa reflected ray)

TOTAL INTERNAL REFLECTIONWhen 2, the angle of refraction becomes 90, the refracted beam is not traveling through the n2 material. Applying Snells law of refraction,

The angle of incidence 1 for which 2 = 90 is called the critical angle c

TOTAL INTERNAL REFLECTION If the ray is incident on the boundary between n1 and n2 materials at the critical angle, the refracted ray will travel along the boundary, never entering the n2 material. There are no refracted rays for the case where 1 c. This condition is known as total internal reflection, which can occur only when light travels from higher refractive index material to lower refractive index material. The light can be restricted to the material with the higher index of refraction if the incident angle is kept above the critical angle. A sandwich of high index material placed between two slabs of low index material will allow a beam of light to propagate in the high index material with relatively little loss. This concept is used in constructing fibers for fiber optic communication.

Index of refraction of cladding is about 1% lower than that of the core.The critical angle is approx 82Core is the center of the optical fiber made of ultra pure glassCRITICAL ANGLEIf light inside an optical fiber strikes the cladding too steeply, the light refracts into the cladding - determined by the critical angle. (There will come a time when, eventually, the angle of refraction reaches 90o and the light is refracted along the boundary between the two materials. The angle of incidence which results in this effect is called the critical angle). n1Sin X=n2Sin90o

Critical AngleCRITICAL ANGLETo propagate light

L: limit angle of propagation

OL: limit angle of acceptance

BASIC PRINCIPLEWhen a light ray travelling in one material hits a different material and reflects back into the original material without any loss of light, total internal reflection is said to occur. Since the core and cladding are constructed from different compositions of glass, theoretically, light entering the core is confined to the boundaries of the core because it reflects back whenever it hits the cladding. For total internal reflection to occur, the index of refraction of the core must be higher than that of the cladding, and the incidence angle is larger than the critical angle.

NUMERICAL APERTURELight must be fed into the end of the fiber to initiate mode propagation. As Figure shows, upon incidence from air (no) to the fiber core (nf) the light is refracted by Snells Law:

The numerical aperture, NA, is defined as

FiberLaser Source

The sum of the internal angles in a triangle is 180 deg.

NUMERICAL APERTUREThe Numerical Aperture is the sine of the largest angle contained within the cone of acceptance.NA is related to a number of important fiber characteristics. It is a measure of the ability of the fiber to gather light at the input end.The higher the NA the tighter (smaller radius) we can have bends in the fiber before loss of light becomes a problem. The higher the NA the more modes we have, Rays can bounce at greater angles and therefore there are more of them. This means that the higher the NA the greater will be the dispersion of this fiber (in the case of MM fiber).Thus higher the NA of SM fiber the higher will be the attenuation of the fiber Typical NA for single-mode fiber is 0.1. For multimode, NA is between 0.2 and 0.3 (usually closer to 0.2).

NUMERICAL APERTURE CALCULATION

ACCEPTANCE ANGLEThe angle of light entering a fiber which follows the critical angle is called the acceptance angle,

= sin-1[(n12-n22)1/2]

n1 = Refractive index of the coren2 = Refractive index of the claddingCritical Angle, cAcceptance Angle,

ACCEPTANCE CONEThere is an imaginary cone of acceptance with an angle The light that enters the fiber at angles within the acceptance cone are guided down the fiber coreAcceptance ConeAcceptance Angle, ACCEPTANCE CONE

SKEW RAYS IN FIBER

ACCEPTANCE ANGLE FOR SKEW RAYS

SKEW RAY PROPOGATION

PROBLEM(1) What happens to the light which approaches the fiber outside of the cone of acceptance? The angle of incidence is 30o as in Fig.1 (calculate the angle of refraction at the air/core interface, r/ critical angle, c/ incident angle at the core/cladding interface, i/) does TIR occur?

PROBLEM(2)Calculate:angle of refraction at the air/core interface, rcritical angle , cincident angle at the core/cladding interface , iWill this light ray propagate down the fiber?nair = 1ncore = 1.46ncladding = 1.43incident = 12

air/core interfacecore/cladding interfaceAnswers:r = 8.2c = 78.4i = 81.8light will propagatePROBLEM(3)Lets find the critical angle within the fiber. Then well find the acceptance angle and the numerical aperture.The critical angle is

The acceptance angle

Finally, the numerical aperture is

PROBLEM(4)A fiber has the following characteristics: n1 = 1.35 (core index) and =2%. Find the N.A and the acceptance angle.n1 = 1.35 ; = 2% = 0.02W.K.T = 1.35 (2 0.02)1/2 = 0.27a = sin 1 (N.A) = sin 1 (0.27) = 15.66Acceptance angle = 2a = 31.33

PROBLEM(5)A silica optical fiber has a core refractive index of 1.50 and a cladding refractive index of 1.47. Determine (i) the critical angle at the core cladding interface, (ii) the N.A for the fiber and (iii) the acceptance angle for the fiber.n1 = 1.50 ; n2 = 1.47 The critical angle

=

The numerical aperture

=The acceptance angle = 2a = 2 sin 1 (N.A) = 2 sin 1 (0.30) = 34.9 Critical angle = 78.5 ; N.A = 0.30 ; Acceptance angle = 34.9PROBLEM(6)Calculate the numerical aperture and acceptance angle of fiber with a core index of 1.52 and a cladding index of 1.50.

Hint: n1 = 1.52 ; n2 = 1.50

= 0.246 and a = sin 1 (N.A) = 1414; Acceptance angle = 2a = 2828

PROBLEM(7)The relative refractive index difference for an optical fiber is 0.05. If the entrance end of the fiber is facing the air medium and refractive index of core is 1.46, estimate the numerical apertureHint:n1 = 1.46 ; = 0.05 ;

RAYSRays & ModesA mode prefers a specific propagation angleAlthough modes and rays behave differentlyFor our purposes: one mode one rayThen, multimode multi-ray

FIBER STRUCTUREWhat is a fiber?Dielectric waveguide (usually cylindrical)An only wireIt does not carry electricityCarries light (inside)Tiny size (like a human hair)Can transmit high data rates

AS FIBER HAS SO MANY PROS, OPTICAL COMMUNICATIONS ARE USUALLY MADE VIA FIBER

FIBER STRUCTUREStructureCore: doped SiO2Carries the light (most of it)Typical values: 8-10; 50; 62.5 m (diameter)Cladding: pure/doped SiO2Confines light into the core (like a mirror)Typical values: 125 m (Coating: outer protection)

FIBER STRUCTUREMain parametersCores size, radius aCladdings refractive index n2Cores refractive index n1 , or n1(r) if it varies

FIBER STRUCTUREFiber types

FIBER STRUCTUREStep-Index Fibern2 = constant (cladding)n1 = constant (core)

Relative Index Difference

Typical values (SiO2): n1 1.461 n2 1.460 SI = 0.01-0.05FIBER STRUCTUREGraded Index Fibern2 = constant (cladding)n1 = n(r) variable! (core)

Relative Index Difference

FIBER STRUCTUREIndex profile parameter g

Most important valuesStep-Index fiber: g Graded index fiber: g 2

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