Download But, with concentration, we may increase this figure by a factor

PROBLEMS APPENDIXES
PROBLEM A1.1 Given that a 0-m star yields an illumination of E0 =
2.65 10-6 lux, calculate the stellar magnitude of the sun and of the moon.
Answ: From Table A1-1, we have that the illuminations supplied by the
sun and of the moon are:
Esun =130 000 lux, Emoon =0.1 lux,
Its stellar magnitude, from Pogson scale, is accordingly:
msun = -2.5 Log10 Esun / E0 + m0
= -2.5 Log10 1.3 105 / 2.65 10-6 = -2.5 (11 –0.3) = -26.7
and
mmoon = -2.5 Log10 Emoon / E0 + m0
= -2.5 Log10 10-1 / 2.65 10-6 = -2.5 (5 –0.4) = -11.5
PROBLEM A1.2 Calculate the loss in dB of the glass of your window
(in the visible).
Answ: Assuming nglass=1.5, at each glass/air interface there is a
reflection loss:
R = [(nglass- nair)/ (nglass- nair)]2= (0.5/2.5) 2= 4%
Transmittance is thus 0.96 and, applying the dB formula (page 363) or
reading directly on the nomogram of Fig.A1-4, we have:
lossdB = 0.17 dB (1 surface), 0.34 (total)
PROBLEM A1.3 For a Si-photodiode, the rated radiant spectral
sensitivity is =0.7 A/W at =850 nm, while the luminous sensitivity is
lum= 5 mA/lm (for a 2850-K lamp). Considering that 1 lm = (1/673)W
= 1.48 mW at =550 nm, how can the two data be reconciled?
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Answ: As the 2850-K lamp used to define the luminous sensitivity has a
rather wide spectral range of emission, the two data are not in contrast
.
By interpolation of the data presented in Fig.A1-5, we can estimate that
the 2850-K blackbody has a spectral emission (at 50% points) extending
from =0.50 to 3.0 m. As the Si-PD can be assumed to have a cutoff at
= 1.0 m, the spectral width of detection is = 500 nm, while it is
V= 100 nm (Fig.A1-3).
Eq. A1.15 gives accordingly (as a first approximation):
lum= (max/Kmax). /V
= 0.7 (A/W) / 673 (lm/W) . 500 nm / 100 nm = 5.2 mA/lm
PROBLEM A2.1 Going out to take a picture of the newly reported comet –
or of the usual night sky - should you prefer to take with you your telescope
(having a 20-cm diameter and a 200-cm focal length) or your camera
(having an objective with 40-mm diameter and a 80-mm focal length) ?
Explain why.
Answ: The photographic film, irrespective of its speed or type (B/N, color)
will have an exposition threshold dictated by the illuminance you are able
to shine on it. Thus, what actually is important, is not just to have the
largest diameter (or collecting power) of the optical system, but its
numerical aperture. The telescope has NA2022000.05while the
camera has NA402800.25 and is therefore better for taking
photographs. Of course, from the standpoint of being able to expose the
faintest stars or light details. The telescope, if its numerical aperture is
sufficient to expose the film, will yield a better angular resolution (stars or
light details better separated).
PROBLEM A2.2. By assuming that sun and planets are blackbodies,
calculate the earth and planets temperature by applying the invariance of
brilliance (it is:Tsun=6000 K). What happens if the emissivity is not unity ?
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Answ: For unitary emissivity, the balance of power collected and emitted
requires that BsunsunAearth= Bsun(sun2)Rearth2= BearthRearth2).
As they are blackbodies, it is Bsun= Tsun4 and Bearth= Tearth4 . Thus, we get:
Tearth4=Tsun4(sun2/4). For Tsun= 6000 K, this yields Tearth= 285 K.
For planets other than earth, Tplanet= 285 K (Learth/Lplanet)1/2 because of the
sun1/Lplanet and Tplanet4sun2 dependences.
If emissivity were not unity, this would leave the sun contribution
unaffected, as its temperature is assumed as the visual one, while planets
would collect less power (a fraction ) from the sun, but also it would
radiate less, just by the same factor
Therefore, provided does not change with wavelength of emission/
radiation, the temperature is the same as above. An exception: if
visible<infrared, then the planet radiates less than that it receives and therefore
the temperature would be considerably higher [by a factor
(visible/infrared)1/4]: this is the case of planet Venus.
PROBLEM A2.3 An engineer of road-lighting crossed his head with the
idea of illuminating tunnels by optical fibers collecting sun light. Using
the brilliance invariance, calculate for him: i) the power that can be
collected from the sun at AM1.5 with a multimode fiber having a 50-m
diameter and a 30°-numerical aperture; and ii) the same, at the
maximum of practicable concentration with a lens (specify which).
Answ: As the fiber acceptance is AW= (/4)D2  (sin 30°)2= 1.5.10-3 mm2,
and the sun gives an illumination E= 130000 lm/m2 at AM1.5, the useable
power is Plum= 130000. 1.5 10-3. 10-6= 0.2 10-3 lm, a very poor starting level
indeed.
But, with concentration, we may increase this figure by a factor
C= (NA) 2/sun2= 50000 (NA) 2, where NA is the numerical aperture of the
collecting lens (i.e., a Fresnel plastic lens with typ. NA=0.5). Also, using a
bundle of N (typ. 1000) individual optical fibers packed together, we can
have Plum= 0.2 10-3 lm . N C= 2500 lm, already a respectable value of some
interest if we can now tolerate the system requirement of adding a
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mechanical movement to the lens to keep the solar disk tracked and focused
on the fiber bundle.
PROBLEM A2.4 Can it be true the old tale of Archimedes devising a
mirror solar-burner of foe-ships sails? Assuming burning temperature of
sail is 300°C, and mirrors held by hands, with 40 cm diameter and 30%
reflectivity (for copper), calculate how many mirrors are required to burn
sails at 200 m distance (of course, assume a sunny day, at noon, no clouds).
Answ: By radiance invariance it is Rsail=Rsun. Writing a power balance of
the received and radiated powers at the sail, we have:
Psail= r Rsun (Amirror / d2) Asail = Tsail4 2Asail,
where the factor 2 is because both sail surfaces radiate, and r denotes
mirror reflectance. Since Rsun=Tsun4 4/ (Lambertian emitter) we get:
Amirror=(2 d2 /r)(Tsail /Tsun) 4 and, inserting numbers,
Amirror=(2 40000 m2 / 0.3) (600/6000) 4 = 83.8 m2.
As the mirror area is ( 0.42m2) /4 = 0.126 m2 we require N=665 mirrors.
PROBLEM A2-5.Consider an Y branching device on a planar substrate,
made of equal lossless waveguides up to the branching region, which on its
turn is smooth and without any geometrical defect. What is the power out of
the lower waveguide of the Y when two equal powers P enter the two upper
waveguides of the Y?
What is the power out of the two upper waveguides when a power P enters
the lower waveguide? What is the phaseshift of the fields E1, E2 out of the
upper waveguides when a field E enter the lower
waveguide (assuming a perfectly symmetrical geometry?)
Answ: By the invariance of radiance, the power out the lower waveguide
cannot be larger than the input power P. Thus, the Y branch has an inherent
and unavoidable 3dB attenuation and it is P = P1=P2.
When exiting at the upper waveguides, the input power P is split evenly
and, because of the structure symmetry, the relative optical phaseshift of the
outputs must be zero. As field vectors E1 and E2 shall add to give E=√P,
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this means that they are both half the amplitude of E, or, in power:
P1=P2=P/4 (once again, there is a 3-dB loss in power).
PROBLEM A2-6.I am not satisfied of the result of Problem A2.5. Add a
more convincing reasoning.
Answ: The results are correct, and the 3-dB loss that undergoes the power
balance can be understood as follows. Consider the mode spatial
distribution in the lower leg arriving at the branching transition of the Ycoupler. We can take it as a Gaussian. When split into the two upper-leg
waveguides, the Gaussian is spatially split in two halves, which are
different from the mode (again a Gaussian) of the guides. Thus, a modeconversion loss is entails, which accounts exactly for the 3-dB loss.
In a X-coupler with 4 input/output ports, no intrinsic loss is found, because
there is no necessary mode conversion. Or, stated differently, the Y-coupler
is a sort of maimed X-coupler, where the missing leg accounts for the loss.
PROBLEM A2.7 To make an endoscope, we want to use a SELFOC fiber
(with parabolic index profile, it re-focalizes periodically the image) to
guide an image of N=500x500 pixel (in visible,=0.5m). If the numerical
aperture is sin=0.3, what is the minimum diameter required?
Answ: By the invariance of acceptance (or of the mode number), we have:
N2 = 0.25.106 2 = A= (/4) D2 (0.3)2
whence we find:
D= 0.5 mm.
PROBLEM A2.8 It is probably well known to the reader that, when laser
light is shed on a diffusing surface, it re-radiated light exhibiting a grainy
appearance called speckle pattern. This is because the diffuser has
destroyed the spatial coherence of the light, at least locally.
54
At a distance z from a diffuser, each speckle-grain has a longitudinal
(parallel to the z-axis) dimension (say sl) and a transversal (perpendicular
to the z-axis) dimension st . By considering that a speckle is an individual
bright (or dark) spot inside which no structure can be resolved, or, it is a
single-mode, calculate its dimensions applying the -acceptance value of
the single mode.
Answ: Consider first the tranversal dimension s t (taken as a diameter) as
seen at a distance z from a diffuser of diameter D.
For a speckle at a distance z from the diffuser, the solid angle within which
rays are coming to the speckle is =(/4)D2/z2, and the receiving surface is
A=(/4) st2. By letting their product equal to 2 for a single mode we have:
A.=(/4) st2.(/4)D2/z2= 2 . By solving for st we get:
st =(4/) zD.
In addition, as the rays are angled of =D/2z around the speckle, they sort
out a length sl =st/ before diverging in size more than st. Thus, the
longitudinal speckle dimension is:
sl =(8/) z2D2.
Worth to note, the near-to-unity multiplying factors of st and sl, namely 4/
and 8/, are dependent on the choice of the illumination on the diffusing
surface. It can be shown that, for a a real laser beam with a Gaussian
spatial-distribution, these factors become exactly unity.
As a last remark, the total volume occupied by the speckle grain, or, by the
coherence region, V=(/4)st2 sl can be computed as V=(8/)  3z4D4 or, for
a Gaussian beam, we have:
V.2= 3,
This equation holds more generally for any single-mode geometry or
situation we may possibly encounter.
PROBLEM A2.9 Should the Greek astronomer Eratostene (200 BC) have
had a photometer, ha could have succeeded in computing the earth-tomoon distance. Hint how.
55
Answ: Eratostene had measured the earth diameter Dearth (looking at the
sun elevation change between Alexandria and Assuan).
He could have proceeded with the following reasoning.
Pointing the photometer to the new moon, he could have determined the
ratio Bm/e/ Bm/s of the brilliance of the moon illuminated by the earth Bm/e
to the brilliance of the moon illuminated by the sun Bm/s . These quantiities
are:
Bm/s=m Bsunsun/, and Bm/e=m Bee/
where Bsun, Be are the sun and earth (surface) brilliance, sun, e are the
solid angle under which sun and earth are seen from the moon and m is the
moon albedo.
Thus, we have: Bm/e/ Bm/s = Bee / Bsunsun.
Now, Be=e Bsunsun/ where the earth albedo e can be measured as the
ratio of the brilliances of a white paper sheet to the (average) earth surface.
Inserting in previous expression yields:
Bm/e/ Bm/s= e Bssune /  Bsunsun.= e e / 
whence the earth-moon distance Le/m is solved as:
e = D2earth / L2e/m = Bm/e/ Bm/s./e or
L2e/m = e D2earth Bm/s /Bm/e
Problem A3.1 At a good illumination level, say ≥100 nit of background
radiance, the eye contrast limit of 0.005 is approached. When you are
under full-lighmoon illumination, what is the limit contrast you can get?
Answ: From Table A1-1, we have that the moonlight illumination is Emoon
=0.1 lux, or Bmin= 1/ Emoon = 0.3 nit.
The Blackwell diagram (Fig.A3-1) gives for such a Bmin value:
Clim = 0.015-0.03 (@ = 200-600 mrad)
or, for a fine detail with = 1.2 mrad, we find the required contrast as:
C1.2mrad = 0.8
Problem A3.2 From data contained in Blackwell’s visual acuity diagram
(Fig.A3-1), estimate the minimum number of photons perceivable to the eye
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in rel time (with an integration time of, say, 0.1 s) in photopic and scotopic
conditions.
Answ: At the boundary (Bmin=2.10-3 nit) of scotopic vision, the Blackwell
diagram can be approximated by the following expression (look at the
Bmin=2.10-3nit point of the second curve from the top):
Bmin.2. C= const. = 2.10-3 . (1.2 mrad)2 . 20 = 57.6.10-9 lux. (1)
For high contrast C it is (see definition in the y-scale of Fig.A3-1):
Bmin C= Bmax- Bmin  Bmax,
the above expression (1) becomes:
Bmax2/4= /4)57.6.10-9 = 0.45.10-7 lux.
On the human eye of normal aperture D=7 mm (in dim light conditions, i.e.
for Bmin=2.10-3 nit), this figure corresponds to a collected power ending in
the single-element sensitive area of the eye:
Bmax2/4(D2/4).
The value of power is:
P= Bmax2 (D2/4)= 0.45.10-7 lux . 38.4 mm2= 17.3 10-13 lm
Now, let us recall that, at =550 nm, it is: 1 lm = 1/673 W , so that
P= (1/673) 17.3 10-13 lm = 25.7 10-16 W.
In T=0.1s, the number of photons received by a power P is:
N = (P/h) T,
and, at 550 nm the photon energy is:
h1.610-19. (1.24/0.55)= 3.6 10-19 W,
whence we get for the minimum number of photons perceived by the eye:
Nphot = (P/h) T = (25.7 10-16/3.6 10-19).0.1 = 710 photons.
On the other hand, in scotopic conditions, which require (as mentioned in
the text) the adaption to darkness for a 10-30 minute period, eq.(1)
becomes:
Bmin.2. C= const. = 3.10-5 . (1.2 mrad)2 . 100 = 4.2.10-10 lux,
that is, we gain a factor 137 in sensitivity respect to photopic vision.
Therefore, the minimum number of photons perceived by the eye is:
Nscot = 710 /137= 5.2 photons
Indeed, this is a remarkable result of performance.
Problem A4.1 Find the variance of the electric field fluctuation for a field
in thermal equilibrium with an ambient at absolute temperature T.
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Answ: From Eq.A4.13, we shall compute
2E = E2 = k /2S/E2E= E

where k is the Boltzmann constant. As it is: Q=TdS=dU-L, we take
dU=0 at equilibrium and express L as:
L= E D. V
where V is the volume of the quantization space,
V= A.L =A c
= Ac /2B in terms of the observation bandwidth B.
Thus entropy is:
S = ∫ Q/T= ∫ - L/T
= ∫ -E D. V/T = -0E2. V/T.
By differentiating twice the above result, we have:
S/E = -20EV/T,
and
2S/T2= -20V/T = -2 A c0 /2BT
Noting that c=1/√00 whence 0 c=√00 =1/Z0 , we get:
2S/T2= 2 A /2BTZ0
so that, by substituting, the field variance is:
2E = k /(A /BTZ0)=1/2 kT (2Z0/A) B
Note the resemblance of this expression to Eq.8.60 derived under the
second quantization condition, where kT and hinterchange.
Problem A4.2 Generalize Eq. 8.60 and determine the wavelength of
transition between the thermal- and quantum-dominated regimes of field
fluctuations at absolute temperature T.
Answ: From the result of Probl.A4.1 and Eq.8.60, we can state that, as a
generalization,
2E =1/2 (kT+h) (2Z0/A) B
Thermal fluctuations of the electric field dominates for kT> h, quantum
fluctuations in the opposite case. The break point is at:
kT > hc or at:
58
 hc kT = 47.9 m
(see also Eq.6.10’).
Problem A5.1 Your 60-MHz oscilloscope has a 1-mV/division vertical
sensitivity. The input resistance is 1-M. Because of Johnson noise
vn= [4kTBR] 1/2, you should then expect to see a trace on the scope with
vn=[4 .40 .10-22 60 .10-6 1 .106] 1/2= 0.98 mV rms noise, that should be
clearly visible. However, no noisy trace is actually seen (trace is much
thinner). How you explain this ? What is the correct expression of your
noise voltage in this case?
Answ: We shall take account of the non-negligible effect of the input
capacitance C across the 1-M resistance. we have also a parasitic
capacitance C, typically C=10 pF. The parallel combination of R and C
yields a high frequency cutoff much smaller than the oscilloscope
measurement bandwidth B. Because of this, the spectral density of noise is
vn2(), =(4kTR) (1+2/RC2), where /RC=1/RC is the high frequency
cutoff of the RC-group. By integrating vn2() from f=0 to f we get for
the total voltage noise seen at the input:
vn2 =(4kTR) ∫0..(1+2/RC2)df = 4kTRRC ∫0..(1+2/RC2)d(/RC)
= 4kTR/2RC atan(/RC) 0..= 4kTR/2RC (/2),
viz. vn2= kT/C.
For example, using C= 10pF for the input capacitance we find:
vn = [kT/C]1/2= [40 .10-22/ 10.10-12] = 0.02 mV
which is well below the scope sensitivity.
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Printed by:
July 2000
Prentice Hall PTR
Upper Saddle River, New Jersey 07458
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