Fiber Optic Communications ( Chapter 2: Optics Review ) presented by Prof. Kwang-Chun Ho 1
Section 2.4: Numerical Aperture Consider an optical receiver: where the diameter of photodetector surface area is d. 2
Numerical Aperture Consider an off-axis beam: Light incident beyond the acceptance angle θ will not be focused onto the detector surface. 3
Numerical Aperture From the previous slide, we see that Then, the numerical aperture is defined as n o is the index of refraction for the surrounding media. It usually has a value of unity. θ is the maximum acceptance angle to detect 4
Numerical Aperture Example: Compute the NA when Solution: Since sinθ tanθ f = 10 cm, d = 1 cm, n o = 1, if θ is so small (i.e. d<<f), we have 5
Numerical Aperture (Applications to Fibers) Fibers have limited acceptance angles and thus a small NA. 6 Typical fibers have an NA in the range 0.1 to 0.5. Coupling to a low-na fiber is more difficult and less efficient than that to a high-na fiber
Numerical Aperture Some NA and the corresponding acceptance angles: 0.5 0.4 0.3 NA 0.2 0.1 7 0 0 5 10 15 20 25 30 Acceptance Angle
Section 2.5: Diffraction Theory Some experiments do not follow ray theory closely enough. A more complete theory based on wave nature of light is needed to explain these ones This is called Diffraction Theory or Physical Optics. As an example of the difference between ray theory and diffraction theory, consider focusing by a lens of a uniform beam of light. 8
Diffraction Theory F. Grimaldi s work in the 1600s! Beam passing a lens Does not converge to A point. Instead, it forms a central spot! Fiber d is the diameter of the central spot. 9
Diffraction Theory The diameter of central spot is found from Example: If f = 2 D, λ = 1µ m, then (2-14) If coupling light into fiber with diameter less than 4 µm, the coupling efficiency will be low. 10
Gaussian Beam Actual light source produce non-uniform beam pattern (Gaussian intensity distribution). 1 0.9 0.8 Normalized Intensity (I/I o ) 0.7 0.6 0.5 0.4 0.3 13.534% 2w where w=10 µm 0.2 0.1 11 0-15 -10-5 0 5 10 15 r
Gaussian Beam (2.5) where w = spot size and r = radial distance from the center of the beam pattern. Note that By definition, w is the spot size. The spot diameter is 2w. 12
Gaussian Beam The Gaussian intensity distribution is useful because most lasers emit in this pattern. And it is the field pattern inside a singlemode fiber. 13
Focusing a Gaussian Beam 2 2w 0 is the spot size of focusing beam appearing in the focal plane 14
Focusing a Gaussian Beam The field pattern in the focal plane is given by 2 2 r I = I e The spot size in the focal plane is given by o w 2 o (2.16) The peak intensity I o will be more intense, because the beam is compressed. 15
Collimating a Gaussian Beam Near the lens the collimation is good (in Fresnel zone), but far from the lens the beam diverges. Fresnel Zone 2w 16
Collimating a Gaussian Beam Far from the lens the beam diverges according to The divergence angle is found from Thus, where (2.17) 17
Collimating a Gaussian Beam Example: Spot size: w= 1 mm at lens, and λ = 0.82 µ m Find θ and w o at z = 10 m, 1 km, and 10 km. Solution: radians 18
Collimating a Gaussian Beam At z = 10 m, 1 km, and 10 km, respectively 19
Beam Divergence Consider beam divergence (2.17) The beam divergence depends upon λ/w. Lenses are analogous to antennas, which direct energy into desired directions. A small beam divergence requires a large spot size relative to the wavelength (λ<<w). 20
Beam Divergence Consider an optical communication system in atmosphere For a long path, much of the light is not collected by the receiving lens. This limits the length of the atmospheric system. Other problems with atmospheric systems are weather and physical interference with the beam. 21
Section 2.6: Summary Review of Subjects presented Refractive index Snell s law Lenses to focus, collimate, and image Numerical aperture Limits due to diffraction theory Gaussian beams 22
P2-8: Problems Compute the divergence angle of Gaussian beam of w= 1 mm and λ = 0.8 µ m If the beam is aimed at moon, what is its spot size on moon s surface? 8 = 3.8 10 m Distance between earth and moon: Solution: zmoon 2λ 4 Divergence angle: θ = = 5.09 10 radians Spot size on moon: π w θ wo == zmoon = 96.7 km 2 23
P2-9: Problems A 6000 km undersea glass-fiber telephone line crosses the Atlantic ocean connecting U.S.A. and France 24