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 1034  3 108
 2.48 1019 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
106

 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 106

 0.522 103 rad
3
w
 10
 z 0.82 106 10
w0 

 2.6 mm
w
 103

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
( f3dB )elec
1/ 2

 0.707
( f3dB )opt 1/ 2
12
( f3 dB )elec 
0.35

or
( f3dB )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
Download

2 - web page for staff

get an essay or any other
homework writing help
for a fair price!
check it here!