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Unformatted text preview: Illuminance (10) Name. Laws of Refraction (10) (Please Print)
D—to—A Converter (20) FI‘IR Spectrometer (20) DIC Microscopy (20) Concave Mirror (20) E C E 4500 First Examination February 14, 2005 Rules for exam: ,ox MAL» 00>] 10.
11.
12.
13. 14. 15. . The time allowed is 60 minutes. The test will start at 9:00am and end at 10:00am. Answer all questions. The value of each question is given in parentheses by the
question. There are a total of 100 points possible. . All work must be shown for full credit.
. Put your ﬁnal answers in the locations speciﬁed.
. You may use the six singlesided 8 l/ " x 11" information sheets that you have prepared.
Some physical constants are given below.
You may use one "book of math tables." . You may use a "pocket calculator" that can be put in a normalsize pocket and requires no external electrical power. Graphing calculators and programmable
calculators are acceptable. You may not use portable, handheld, laptop, or
notebook computers or wireless or network connections. . You may not use any reference materials other than those listed above. Therefore, you may not use the class notes, any textbooks, homework problems, reprints
of papers, journals, prayer books, etc. There is to be no sharing of anything. If excess information is given in a question, ignore the unneeded information. If too little information is given in a question, assume the information needed and
clearly note this with your work. Any changes to the examination will be written on the board. Check the board
periodically during the examination. Any acts of dishonesty will be referred to the Dean of Students without prior
discussion. The ofﬁcial written Institute procedures on academic honesty (entitled
"Maintaining Academic Honesty" and available from the Dean of Students Ofﬁce)
will be followed in all cases. Have a happy exam! = 6.6260755 x 10'34joulesec
= 299792458 x 108 meter/ sec
= 1.6021773349 x 10'19cou1 k = 1.38065812 x 10'23joule / K 005‘ Table 1 Susan tumour! DATA WW Lam(M ‘VMM MW CM Illuminance It is desired to measure the illuminance at the screen of a frontprojection high—
deﬁnition television (HDTV) system. Using one complete sentence, state the deﬁnition
of “illuminance” for this situation. Give two possible units for illuminance. Indicate the V , appropriateinstrument for measuring illuminance. Put your answers in the spaces pro— vided. Deﬁnition of Illuminance: Two Possible Units of Illuminance: Illuminance Measuring Instrument: (circle the one best answer) (spectrometer) (interferometer) (radiometer) (photometer) (refractometer) (microscope) (telescope) (demultiplexer) Laws of Refraction As described in the course notes and in the textbook, there are two laws of re
fraction. The ﬁrst law states that the refracted wave is phase matched to the incident
wave. For a wave in a dielectric of index n1 incident uponan interface with dielectric of
index 712, the angle of refraction 82 is given by the relationship n1 sin 01 = mg sin 02. For
isotropic materials this is Snell’s law. Using one or more complete sentences, state the second law of refraction. Using
additional complete sentences, deﬁne all quantities that you have used in your statement
of the second law of refraction. Put your answers in the spaces provided. Statement of Second Law of Refraction: Deﬁniti0n(s) of All Quantity(ies): ElectraOptic DigitaltoAnalog Converter Segmented interdigitated electrodes are deposited on top of a lithium niobate slab
waveguide as shown. The diffraction efﬁciency (fraction of the incident power that is
diffracted) for each segment is given by 77 = sin2(AV), where V is the voltage applied to a. grating segment and A is a constant. With four
grating segments, this device is to be used to convert an electronic 4bit binary word
into an optical analog value as represented by the total power of the optical wave being
focused onto the detector. The voltage for each bit is “on” for a 1 bit and “off” for a
zero bit. Any segment of the interdigitated electrodes can have a diffraction efﬁciency of
100% by applying a voltage V 2 VB to that set of electrodes. The difference in the pow
ers produced by the A—to—D converter for the binary numbers 0000 and 1111 should be
as large as possible. ElectricalDigital Input 20 21 22 23 OpticalAnalog
Output Uniform
Guided
Wave Calculate, showing all work, the voltage for each grating segment required for
each bit of the A—toD converter. Express your answers as fractions of V0. voltage for least signiﬁcant bit, V0001 2 V0.
voltage for next signiﬁcant bit, V0010 2 V0.
voltage for next signiﬁcant bit, V0100 = Vb.
voltage for most signiﬁcant bit, V1000 = Vb. Fourier Transform Infrared Spectrometer A Fourier Transform Infrared (FTIR) spectrometer is based on a Michelson in
terferometer. An FTIR spectrometer is capable of determining the spectral content of
a light source. This is accomplished by scanning one of the mirrors of the Michelson in
terferometer and recording the output interferogram as a function of the optical path
difference between the two interfering beams. In an FTIR, the wavelength of a wave is represented by its wavenumber v where l
A U: and /\ is the freespace wavelength. A Fourier transform relationship exists between the
spectrum in Wavenumber space ('0) and the amplitude interference pattern (interfero gram) in optical path difference space (6). Interferograms obtained for two sources are
shown in the attached ﬁgure. The interferogram for the ﬁrst source represented by the upper ﬁgure is
S1(6) = A1 cos(27rv1 6) + A2 cos(27rv2 6).
The interferogram for the second source represented by the lower ﬁgure is 82(5) '2 A3 605(271'01 6) + A4 008(27T’Ug 6). From these two source amplitude interferograms, answer the questions given below. For S 1 (6) (circle the one best answer) (lAll = lAzl) (A1 = A2) (lAll # [A2) (none of these) For 5'1 (6), estimate, showing all work, the ratio 112/121. Put your answer in the space pro
vided. ‘ 02/111 = For 82(6) (circle the one best answer) (IA3l : A4) (A3 = A4) (lAal 5‘ A4l) (“0116 Of these) For these two interferograms, draw the corresponding amplitude spectra in the space provided. AMPUTUDﬁ‘ AMPUTUDE O 1©N1
OFTEAL WWH DWFERENCE AMPLHUDE AMPUTUDE WAVENUMBER Differential Interference Contrast (Nomarski) Microscopy 2 A Differential Interference Contrast (DIC) or Nomarski microscopy conﬁguration is shown in the diagram. Analyzer I:::I Beam
Combiner Objective Channel
Wavguide Substrate Condenser Beam
Splitter Polarizer :3 Light of freespace wavelength A is used in this microscope. The modiﬁed Wollaston
beam splitter produces two orthogonally polarized beams (solid line and dashed line)
that have a lateral displacement (shear) that is less than the resolution of the objective.
This DIC microscope is used to examine a channel waveguide as shown in the ﬁgure.
The waveguide material is a polymer with an index of refraction of nu, at wavelength A.
The width of the waveguide is Wu, and the thickness of the waveguide is Lu, as shown.
The polymer is deposited on a substrate that has a refractive index of ms at wavelength
A. The thickness of the substrate is L s as shown. The waveguide and substrate are sur—
rounded by air. In the ﬁgure, the two orthogonally polarized beams are shown at both the left edge of the channel waveguide and at the right edge of the channel waveguide. When the above channel waveguide is vieWed through the DIC microsCope, the beam
combiner is adjusted so that the phase retardation bias introduced produces the max—
imum transmittance (or intensity) contrast between the left edge of the channel wave—
guide and the right edge of the waveguide. Under these conditions, derive, showing all
work, the transmittance diﬂerence, AT, between the left edge and the right edge of the
waveguide. Express your answer only as a ftmction' of the parameters given above. Sim plify your answer as much as possible. Put your ﬁnal answer in the space provided. Concave Mirror 2 A concave spherical mirror is in air as shown in the attached ﬁgure. The magni
tude of the radius of curvature of the mirror is 100 mm. A vertical planar object exists
75 mm to the left of the mirror. A point on the object is located 50 mm above the axis of the optical system. Calculate, showing all work, the focal length (including sign) of the mirror. Cal—
culate, showing all work, the location of the image along the optical axis to the left or
to the right of the mirror. CaICulate, showing all work, the location of the image point
above or below the optical axis. Express your answers in mm. For the image distance,
indicate whether it is left or right of the mirror. For the image point, indicate whether it is above or below the optical axis. Put your answers in the spaces provided. In the attached ﬁgure, the spacing between tick marks is 50 mm. On the drawing,
put a dot at the center of curvature and at the focal point. On the ﬁgure, draw a ray
trace diagram that incorporates 1) the parallel ray, 2) the chief ray, and 3) the focal ray.
Label the parallel ray, the chief ray, and the focal ray. 1 Focal length 2 mm
(Sign)
Object Image
Distance Distance Distance
along axis left or right along axis left or right from axis above or below 75 left Illuminance The “illuminance” is the luminous power (or luminous ﬂux) per unit area impinging
on the screen. Possible units of illuminance are lumens/m2, footcandles, phat, or lust. A photometer is an illuminance measuring instrument. Laws of Refraction The second law of refraction states that the refracted wavevector lies in the plane of
incidence. ' The refacted wavevector is a vector of magnitude of 2 7r 112 /A Whose direction is the
direction of phase propagation of the refracted wave. The quantity A is the freespace
wavelength. ' The plane of incidence is the plane deﬁned by the incident wavevector and the normal
to the interface where the incident wavevector strikes theinterface. ElectroOptic DigitaltoAnalog Converter
Bragg regime
1} = sin2 (AV) 01' V = Zsin‘lﬂ = 271'
and so V =%(sin’1\/ﬁ)Vb Therefore, V0001 = g— [Mﬂﬁﬂ V0 = 02301 V2, for n = 1/8.
V0010 = %Vb = 0.3333Vo for n = 1/4. V0100 = %Vb = 0.500014; for n = 1/2. V1000 = V0 for 77 = 1 Fourier Transform Infrared Spectrometer The ﬁrst interferog‘ram is
5'1 (6) = A1 cos(2 7r «)1 6) + A2 003(2 7r 02 6). From the ﬁrst interferogram, it is observed that when the two waves are out of phase (at
5/ 121) they cancel completely. Thus From the interferogram, it is observed that the waves are in phase at 6 = 0. As the path
difference increases, the waves are 180° out of phase at 6 2 5/121 (5A1 wavelengths) and
then are back in phase at 6 = 10/ 01 (10M wavelengths): There are a total of about 11
amplitude peaks for 0 < 6 < 10/ 111. If A1 alone were present there would be 10 ampli
tude peaks. Therefore the seocnd wavelength A2 must be such that 11 A2 = 10 /\1. Thus A1 112 11 A—z ‘Ul E
The second interferogram is
32(6) 2 A3 cos(2 7r v1 6) + A4 003(2 7r v3 6). From the second interferogram, it is observed that when the two waves are out of phase (at about 6 2 22/121) they do not cancel completely. Thus lAal 7é lA4l ‘ For these two interferograms, the amplitude spectra are shown. AMPUTUDE AMPUTUDE 0 . 1©N1
OPTEAL WWH DWFERENCE AMPUTUDE AMPUTUDE W‘Q V1V3
WAVENUMBER Differential Interference Contrast (Nomarski) Microscopy 2 freespace wavelength = A channel waveguide refractive index channel waveguide thickness =Lw left edge of channel waveguide (An)1 right edge of channel waveguide (An)2 T = II": — sinz—g—
where
I‘ = I‘mas + 277rAnL For maximum contrast Fbias : 77/2
andso
7r 27r
I‘ = — —A L
2 + A "
1r 27r
I‘ = — — L
1 2 + )‘(A’nh
7r 271'
I‘ = — — A L
2 2 + A( “)2
T1 — sin2[ 2
. I“; +2—’r
T2 = szn2[ “8 3 ATle—Tzzsin This can be simpliﬁed using cos 20 1 1
«2 9____
3m 2 2 ”111 nw—l = l—nw 7r 7T 4 A ’7? 7T 4 /\ (nu, —— 1) Lw] — sin2 [ sin2 [— + — (nu, — 1) Lw] sinz [— + — (l — nw) Lw] 3+”
4 /\ Rood; + k(nw—1)Lw]  0045 + k(1‘"w)L‘"]) AT 2 2 where k = 27r/A. This can be further simpliﬁed using cod; + 0) = —sin0
1 . .
AT = 5 (sm [k (1 — nw) Lw] —— 3m [k (nu, — 1) Lw])
Using
sin(—0) =  sin0 AT 2 sin [k (1 — nw) Lw] If the arguments are sufﬁciently small, this could be further simpliﬁed using the small argument "approximation sin 0 = 0. Concave Mirror 2 Mirror parameters R 2 —100mm, 3 = 75mm f = e; = +50mm l+i__3 s s’ _ R.
1 3R I: =  = 150 1ft f ' s _%_l 2s+R + mm (e omirror)
s/ m = — g = —2.000 (inverted real image) Distance from axis I 2 m0 = (—2) (50mm) 2 —100mm (below axis) ...
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 Spring '08
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