ENE 423 Lecture II Optics Review Lights are made up of very small particles called photons. As we know, the energy of a photon is E hf where h = 6.626 x 10-34 J/s (Planck’s constant). This assumption could explain large-scale optical effects such as reflection and refraction, but cannot describe finer-scale phenomena such as interference and diffraction. The correct explanation of diffraction was suggested by Fresnel. He showed that light is a wave motion. Later, Maxwell said that light is electromagnetic in nature. Ex. Find the number of photons incident on a detector in 1 second if the optic power is 1 μW and the wavelength is 0.8 μW. Soln E hf From P hc 6.626 1034 3 108 2.48 1019 J 6 0.8 10 E , this implies that the energy 1 J per 1 s yields the power of 1 W. t Therefore, E 106 4.03 1012 photons 19 E p 2.48 10 Optical energy is assumed to be confined and travel in narrow paths called rays. Rays obey a few simple rules as 1. In air, rays travel at a velocity of c. In any other medium rays travel at a slower speed, given by v = c/n where n = refractive index of medium (air: n=1, water: n=1.33, fiber or glass: n=1.45-1.5, diamond: n = 2.42) 2 2. Rays travel in straight paths unless deflected by some change in the medium. 3. At a plane boundary between 2 media, a ray is reflected at an angle equal to the angle of incidence. θr θt θi θi = θr n1sin θi = n2sin θt Snell’s Law: In case of θi = 0, then sinθt = 0. Snell’s law yields sinθt = 0 or θt = 0. This implies that the ray is not deflected. air: n1 = 1 and fiber: n2 1.5 θt From Snell’s law: θt < θi θi Ex. A light ray travels from air into glass (n = 1.5). Find the transmission angle when θi = 15. Then it travels from glass back into air. Find the direction of the transmitted ray. Soln n1 sin i n2 sin t sin15 9.94 1.5 From glass back to air:air 15 sin t This shows that a ray incident on a parallel plate of glass will suffer no net deflection (deflected back to its original direction). 3 Lenses Lens is used for coupling light in and out of optical components. Focusing a light beam onto a fiber. Collimating a diverging light beam. Normally, a lens is constructed by connecting the caps of two solid glasses. Therefore, it has two curvatures of R1 and R2. The focal length of a lens can be calculated by 1 1 1 n 1 f R1 R2 If the radius of the sphere is the lens curvature, and the lens diameter is twice the curvature or R1 = R2 = D/2, this yields f D 4 n 1 The ratio of f/D is called “f-number” of the lens where f is a lens focus length and D is a lens diameter. Imaging An image can be formed by a thin lens as shown in the figure below. The thin lens equation used to find the position of an image can be written as 1 1 1 d0 di f 4 The magnification M of the lens could be expressed by M di d0 Ex. Find the object and image distances if M = 1. Soln M 1 yields di d 0 1 1 1 d 0 di f d0 2 f Gaussian Beam Actual light sources often produce non-uniform beams. The intensities vary across the transverse plane. This intensity distribution has a Gaussian distribution that is given by I I0 e 2r2 w2 5 An accepted definition of the radius of the spot is the distance at which the beam intensity has dropped to 1/e2 = 0.135 I0. This radius (r = w) can be called “spot size”. Focusing Gaussian beam with a lens yield a distribution of light in the focal plane that is also Gaussian shaped. w0 f w Collimating a Gaussian beam with source located at the lens focal point, this predicts a parallel beam will emerge from the lens. For longer distances, the beam will diverge at a constant angle. 2 w z w0 w 6 Ex. Consider a Gaussian beam whose spot size is 1 mm when collimated. The wavelength is 0.82 μm. Compute the divergence angle and the spot size at 10 m. Soln The divergence angle is 2 2 0.82 106 0.522 103 rad 3 w 10 z 0.82 106 10 w0 2.6 mm w 103 Lightwave Fundamentals Light consists of an electric field and a magnetic field that oscillate at very high frequency of 1014 Hz. In optics, an electric field is often used to specify lightwave. Assume that light is traveling in a homogeneous isotropic and lossy medium and it is a linearly light propagating in z-direction. The electric field can be expressed as E E0 e z sin(t kz ) where E0 = peak amplitude = attenuation coefficient (loss) ω = 2πf = angular frequency of light k = propagation constant = ω/v v = phase velocity The factor ωt – kz is the phase of the wave and kz is the phase shift due to travel over length z. The free-space (air) wavelength is 0 any medium is c and the wavelength in f c v , so that 0 n . v f The light intensity is proportional to the square of its electric field. Therefore, the power reduction in decibels for a length L can be expressed as dB 10log10 exp 2 L 7 Dispersion Dispersion refers to velocity dependence on and causes signal distortion caused by material dispersion and/or waveguide dispersion. It is important because light radiates from source over a range of s known as spectral distribution. The dispersion may be categorized as material dispersion and waveguide dispersion. Source Linewidth or spectral width (nm) Light-emitting didoe 20 - 100 Laser diode 1-5 Nd:YAG laser 0.1 HeNe laser 0.002 When light is modulated by a signal, then different s within travel at different velocity because v c / n( ) . Therefore, the pulses are reaching the end of the fiber at different times. The further the pulse travels, the greater the spreading. 8 Digital Analog d 2n l. . The pulse spread is given by 2 c d Then the pulse spread per unit length . l l 9 l 2 d n . M . 2 c d d 2n M . 2 c d M is defined as the material dispersion, generally in unit of ps/(nm.km). Material dispersion for pure silica If we let a travel time be 1 for 1 and 2 for 2, then = (2 - 1). If > 0 or 2 > 1, this implies 1 travels faster than 2. If < 0 or 2 < 1, this implies 2 travels faster than 1. The sign () is important if both material and waveguide dispersion are taken into account. For the figure above, 0 for M = 0 could be shifted by doping fiber with some materials. A useful approximation for values being near 0 is M M0 04 4 3 where M0 = the slope of (M-) curve over the range of 1200 nm < < 1600 nm = -0.095 ps/(nm2/km) To minimize distortion use - source with near 0 and small - solitons (pulse propagates without spreading.) It is based on non-linear effect that requires high optical power. 10 Ex. Find the amount of pulse spread in a 10 km pure silica fiber when the light source is (a) LED at = 0.82 micron of = 20 nm (b) LED at = 1.5 micron of = 50 nm (c) LD at = 0.82 micron of = 1 nm Soln (a) From the figure, M = 110 ps/(nm.km) at = 0.82 μm = -[110 ps/(nm.km)][10 km][20 nm] = -22,000 ps (b) From the figure, M = -15 ps/(nm.km) at = 1.5 μm = -[-15 ps/(nm.km)][10 km][50 nm] = 7,500 ps = 7.5 ns (c) From the figure, M = 110 ps/(nm.km) at = 0.82 μm = -[110 ps/(nm.km)][10 km][1 nm] = -1,100 ps = -1.1 ns **We clearly see from (a) & (c) that the narrower the spectral width (), the less the distortion. Dispersion Effects on Information Rate Analog system Consider a sinusoidal modulated signal of light with modulated frequency f and the period T (T =1/f). Suppose source radiates 1 and 2. At receiver, when the delay is equal to half of the period, the modulation cancels out completely. 11 If a maximum allowable pulse spread is equal to T/2, then the modulation frequency is limited by f 1 1 T 2 This f provides a good approximation to the 3-dB bandwidth (f at which output optical signal intensity drops by half). Optical power, P0(f) i(t) P0(f) RL Optical case: P0 ( f ) 0.5 P0 (dc) Electrical case: P0 ( f ) 0.707 P0 (dc) Note: electrical power (optical power)2 and light intensity 1/f This leads to ( f3dB )elec 1/ 2 0.707 ( f3dB )opt 1/ 2 12 ( f3 dB )elec 0.35 or ( f3dB )elec l 0.35 ( / l ) Digital System - Return to zero (RZ) Most of received signal power is contained within 1/T and this 1/T implies a required bandwidth for the system. Data rate R equals to 1/(bit period). RRZ 1 0.35 ( f3 dB )elec T RRZ l 0.35 ( / l ) 13 - Non return to zero (NRZ) Most of received signal power is lied within 1/(2T) which is a bandwidth requirement. Approximating sinusoidal has period 2T so f = 1/(2T). RNRZ 1 0.7 2( f3 dB )elec T RNRZ l 0.7 ( / l ) Ex. Data are transmitted through fiber at rate 48 Mb/s using RZ format. Fiber has a length of 76 km, M = -20 ps(nm.km). Find required spectral width of light source. Soln RRZ = 0.35/, = 0.35/(48 ×106 b/s) = -M.l. 7.29 x 10-9 = (20 × 1012 s.km-1.nm-1)(76 km)( ) = 4.78 nm

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