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Unformatted text preview: General Questions (18) ' Name Slab Waveguide (20) (Please Print)
Interference Filter (22) Circular Aperture (16) Transmittances (24) E C E 4500
Third Hour Examination
April 27, 2004 Rules for exam: 1.. 2. 3
4.
5 0°99 10.
11.
12.
13. 14. 15. The time allowed is 60 minutes. The examination will start at 2:50pm.
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 six new singlesided 8'1/ " x 11" information sheets that youhave prepared. You may also use the six singlesided information sheets that you
prepared for the ﬁrst examination as well as the six single—sided information sheets
that you prepared for the second examination. Some physical constants are given below. You may use one "book of math tables" or calculus textbook. . 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, infrared, 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
16021773349 x 10‘19 coul k = 138065812 x 10'23joule / K (905' General Optical Engineering Questions For each question below, circle the one best answer. 1. At a freespace wavelength of 600 nm, the refractive index of gold is approximately
a) 1
b) 1.5 f) 0.5 — 3'3
g) all of the above 2. Elliptically polarized light of freespace wavelength A can be converted to linearly po
larized light by a a) quarter waveplate b) half waveplate c) full waveplate d) an interference ﬁlter e) a parabolic mirror f) none of the above 3. At a freespace wavelength A, the TED mode is most tightly bound in the slab waveg—
uide (where us is the cover refractive index, nf is the film refractive index, and n3 is the substrate refractive index) described by
a)nc=1,nf=1.5,n5=1 b) nc=2,nf=3, 113:2 0) nc = 1.5, nf = 2.5, ns = l5
d)nc=1,ﬁf=2,ns='_—1 e) none of the above 4. A thin slab of thickness d and refractive index n with ﬂat and parallel surfaces will transmit normally incident light of freespace wavelength A for m being an integer if
a) 2nd 2 mA b) 2nd —._ (m — ax c) nd 2 mA d) and z (m — a A e) 4nd 3 mA f) 4nd 2 (m — %),\
g) none of the above 5. For a dielectric substrate of refractive index n‘.J in air, for a single—layer coating to be antireﬂecting, it should have a refractive index of f) ns+1 g) (“a ‘l' 1)/2
h) none of the above 6. A reﬂection hologram is recorded when two mutually coherent beams are incident on
(the same side) (opposite sides) of a high—resolution photographic recording plate. (can— not be determined from given information) (none of the above) Single TEo Mode Slab Waveguide A singlemode titaniumdiffused lithium niobate slab waveguide (nf = 2.234) is to
be constructed for telecommunications applications. It is surrounded by a lithium nio—
bate substrate (ns = 2.214) and an air cover (nc = 1.000). Light of freespace wavelength 100pm is to be guided in. the waveguide. Calculate, showing all work, the range of thicknesses, h, for which only a sin
gle TEO mode will propagate. The TMO mode, the T191 mode, the TMl mode and all
higherorder modes are to be cut oil. Express the minimum and maximum thicknesses in um accurately to four signiﬁcant ﬁgures. um<h< pm Wavelength Selection with an Interference Filter A “one half wavelength” thick interference ﬁlter is passes blue light from an argon
laser of freespace wavelength 488.0 nm when it is rotated so that the angle of incidence
in air is or = 30°. This is shown in the diagram below. The interference ﬁlter is com—
posed of parallel layers of refractive indices 1.5 and 1.6 as shown in the ﬁgure. In a sec—
ond angular orientation, it is desired that the interference ﬁlter pass green light from an
argon laser of freespace wavelength 514.5nm. In a. third angular orientation, it is desired that the interference ﬁlter pass violet light from an argon laser of freeSpace wavelength
476.5 nm. Calculate, showing all work, whether or not it is possible for this given ﬁlter to
pass 514.5 nm light and whether or not it is possible to pass 476.5 nm light. If either
or both of these cases are possible, calculate, showing all work, the required angles of
incidence, a, to accomplish them. Put your answers in the spaces provided. If angles of
incidence are realizable to pass these wavelengths, express these angles accurately to four signiﬁcant ﬁgures. Is it possible for this ﬁlter to pass 514.5 nm light? (yes) (no)
(cannot be determined from information given) (Circle one) _ If it is possible, then a (514.5.nm) 2 ° Is it possible for this ﬁlter to pass 476.5 nm light? (yes) (no)
(cannot be determined from information given) (Circle one) If it is possible, then a (476.5nm) = ‘ ° Eraunhofer Diffraction by a Circular Aperture The characteristic Eraunhofer diffraction pattern due to a circular aperture illu— minated at normal incidence is known as an Airy disk and the radiant intensity is given
by
JﬂstinQ/A) (11’ D sin 9/ M2 where J1 is the ﬁrst—order ordinary Bessel function of the first kind, D is the diameter 1(9) = 410 of the aperture, and 6 is the angle measured from the center of the aperture to an arbi
trarg,r point in the diffraction pattern. A circular aperture is shown at left in the diagram below. The resulting farﬁeld Airy disk pattern is shown at right in the diagram below. l A
___...... Lx
—— {ML l~————Lz——i For the particular case of a circular aperture of 200 nm diameter illuminated by
a helium—neon laser of freespace wavelength 632.8 nm, calculate the diameter L:c of the
ﬁrst dark ring in the Airy disk pattern when the observation plane (shown at the right
in the ﬁgure) is L2 = 10 meters. Express your answer in mm accurately to four signiﬁ cant ﬁgures. Put your answer in the space provided. L3 2 mm FarField Diffraction by Rectangular and Triangular 'Iransmittance Slits A rectangular transmittance slit and a triangular transmittance slit have ampli
tude transmittances as shown below. transmittance The rectangular slit has a width a and a peak amplitude transmittance of 0.5.
It is illuminated in air with a collimated laser beam of freespace wavelength A. In the
farﬁeld diffraction pattern, a bright central peak surrounded on both sides by nulls and
a series of dimmer peaks is observed. The bright central peak is taken to be at 6 = 0
where 9 is the angle measured from the center of the slit to the far ﬁeld. For this diHrac
tion pattern, write down (it is not necessary to derive) the expression for the farﬁeld
radiant intensity I Express the radiant intensity as a function of a, 9, A, and I0 only
where I0 is the value of I (0) at 6 = 0. Develop the exact expression for the angle in ra
dians from the central peak to the ﬁrst null (61%”). Express the angle as a function of a
and A only. Put your ﬁnal answers in the spaces provided. Rectangular slit radiant intensity I (6) = For rectangular slit, farﬁeld angle from central peak to ﬁrst null = radians The triangular slit has a width (1 and a peak amplitude transmittance of 0.5. It is
illuminated in air with the same collimated laser beam of freespace wavelength A. In the
far—ﬁeld diffraction pattern, a bright central peak surrounded on both sides by nulls and
a series of dimmer peaks is observed. The bright central peak is taken to be at 6 = 0
where 6 is the angle measured from the center of the slit to the far ﬁeld. For this diffrac
tion pattern, write down (it is not necessary to derive) the expression for the far—ﬁeld
radiant intensity I Express the radiant intensity as a function of a, 6, A, and In only
where I0 is the value of 1(6) at 6 = 0. Develop the exact expression for the angle in ra
dians from the central peak to the ﬁrst null (9p_n). Express the angle as a function of a
and A only. Put your ﬁnal answers in the spaces provided. Triangular slit radiant intensity I (0) = For triangular slit, farﬁeld angle from central peak to ﬁrst null = radians General Optical Engineering Questions For each question below, circle the one best answer. 1. At a freespace wavelength of 600nm, the refractive index of gold is approximately
f) 0.5 — 3'3 2. Elliptically polarized light of freespace wavelength A can be converted to linearly po larized light by a
a) quarter waveplate 3. At a freespace wavelength A, the TED mode is most tightly bound in the slab waveg
uide (where nC is the cover refractive index, nf is the ﬁlm refractive index, and 11,3 is the substrate refractive index) described by
d) nc=1,nf=2,n3=1 4. A thin slab of thickness d and refractive index n with flat and parallel surfaces will transmit normally incident light of freespace wavelength A if
a) 2nd = m A 5. For a dielectric substrate of refractive index n, in air, a singlelayer antireﬂection coating should have a refractive index of
d) iii/2 6. A reflection hologram is recorded when two mutually coherent beams are incident on (opposite sides) of a highresolution photographic recording plate. Single TED Mode Slab Waveguide nc = 1.000 nf = 2.234 125 = 2.214 A=1.00,um
'2_ 2 em = “g "’3 = 43.8601
nf_n5
“4 2_ 2 am — —ﬁ "3 n; = 1092.45
nc nf—ns VCOTEO = ten—w—em = 81.4130 = 1.42093md v.30”,o = tanV—“em = 88.2670 = 1.54055md A WOTE A tan—lanE
heoTEo = W = W = 0.75823 ,um
_ AVCOTMD _ Atan'dwaTM _
hcoTMo — W — W — 0.82205pm and so for true Singlemode (TED) behavior, the thickness should be 0.75823pm < h < 0.82205pm Wavelength Selection with an Interference Filter Angle of incidence a1 = 30° passes /\1 = 488.0 nm
Parallel glass plates do not affect the ray direction in air. Thus from Snell’s law
1.0 sin (11 = 1.5 sin [31 and so ﬁl = 1947122". The angular tuning is described by
2 n (1 cos [5’ = m A For one half wavelength ﬁlters at normal incidence
2nd = A” and so /\ ‘ A
cos [31 = A—1 and thus An = wslﬂl
n 2 517.6021 nm at normal incidence For tuning to A2 = 514.5 nm A
COSﬁg = A—2 and so [32 = 6.276080 1.03ina2 = 1.5sinﬁg and so 052 = 943794". For tuning to A3 = 476.5 nm % and so {33 = 22.987330 11. II 005 153 1.03ina3 = 1.53inﬁg and so 053 2 3585892". Fraunhofer Diffraction by a Circular Aperture The Rayleigh criterion
A9p_n = 1.22% gives the angular distance from the intensity peak to the ﬁrst null. The angular distance nullto—null is
M“, = 2.44% = 0.007720md = 0.44233°.
The nulltonull linear distance is L,._n = L3, = Lz A6n_n = (10m)(0.007720rad) = 77.2mm FarField Diffraction by Rectangular and Triangular Transmittance Slits Rectangular slit sin2 (E32)2 F2(jkma:) = (12 Since kg; = ZT‘TSinﬂ and I0 = a2 SinZ («11:11:10 )
( 1man ) 2 I(6) = ID The ﬁrst null occurs at ﬁasinﬂ _
A — ﬂ 9 = sin—1e) Triangular slit 2 maﬁa) _ (1
Max) = I k a 4
‘11) Since kg: = 271752716 and I0 = (12/4 Sin4(1mzsin6) nasinﬂ ) 4
2A 1(9) = I0 The ﬁrst null occurs at wasin 9 _
2 A _ 7T 9 2 sin—1(2 ...
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This note was uploaded on 04/29/2008 for the course ECE 4500 taught by Professor Gaylord during the Spring '08 term at Georgia Tech.
 Spring '08
 Gaylord

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