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Unformatted text preview: General Questions Two Lenses
Thick Lens Imaging ( 15) (21) Name (20) (Please Print) Microscope Objective (12) Hemispherical Lens
Metal Reﬂection (18) (14) E C E 4500 Second Examination April 2, 2003 Rules for exam: 1. SAPS” 9°99 10.
11.
12.
13. 14. 15. The time allowed is 60 minutes. The test will start at 9:00am (rather than at
9:05am) and end at 10:00am (rather than at 9:55am).
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 you have
prepared. You may also use the six singlesided information sheets that you
prepared for the ﬁrst 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, 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 chalk board. Check the
chalk 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‘34joule—sec
2.99792458 x 108 meter / sec
1.6021773349 >< 10‘19coul = 138065812 >< 10'23joule / K h
c
e
k General Optical Engineering Questions 1. A light ray is incident at an oblique (nonnormal) angle on a boundary between di
electrics. The incident light ray is propagating in a relatively low refractive index dielec
tric. On the other side of the boundary is a relatively high refractive index dielectric. The ray refracted into the relatively high refractive index dielectric is (Circle one.)
a) deviated toward the normal to the boundary b) undeviated relative to the normal boundary c) deviated away from the normal to the boundary d) none of the above 2. The linear magniﬁcation (or transverse magniﬁcation or lateral magniﬁcation) of an
optical system is m, where m = s’/s and s is the object distance and s’ is the im—
age distance. The longitudinal magniﬁcation, on the other hand, is As’/As. Write an
expression for the longitudinal magniﬁcation as a function of m. The longitudinal mag—
niﬁcation As’/As = 3. Of the aberrations coma and astigmatism, which is more important to be corrected in an astronomical telescope? (Circle one.)
a) coma b) astigmatism c) none of the above 4. Linearly polarized light is propagating in glass of refractive index of 1.50. This light
is incident normally upon plane boundary with air (refractive index of 1.00). Immedi
ately upon reflection at the boundary, the electric ﬁeld of the electromagnetic wave has a phase shift (relative to the phase of the incident wave) of (Circle one.)
a) 0 degrees b) 180 degrees c) 0 < phase shift < 180° d) none of the above 5. A TE polarized light wave is propagating in a dielectric of refractive index of m. This light is incident upon plane boundary with a second dielectric of refractive index 112 (m > 712). The angle of incidence is greater than the critical angle. Immediately upon
reﬂection at the boundary, the electric ﬁeld of the electromagnetic wave has a phase shift (relative to the phase of the incident wave) of (Circle one.)
a) 0 degrees b) 180 degrees c) 0 < phase shift < 180° d) none of the above 6. Circularly polarized light of freespace wavelength A can be converted to linearly po— larized light by which of the following waveplates designed for that wavelength. (Circle one.) a) zero waveplate b) quarter waveplate
0) half waveplate (1) full waveplate e) none of the above 7. For A being the freespace wavelength, in a typical glass of refractive index n in the visible portion of the electromagnetic spectrum (Circle one.)
a) dn/dA < 0 b) dn/dA = 0 c) dn/dA > 0 d) none of the above Imaging with Two Lenses Two thin lenses in air are used to image an object. Lens no. 1 is a converging lens
with a focal length of f1 2 +100 mm. Lens no. 2 is also a converging lens. Lens no. 2
has a focal length of f2 = +50 mm. The object is 150 mm to the left of lens no. 1 as
shown in the ﬁgure. The distance between tick marks in the ﬁgure is 50 mm. The dis tance between these two thin lenses is 200 mm. Lens Lens
No. 1 No. 2
Lens
No. 1
Lens
No. 2 (continued on next page) Calculate, showing all work, the location and magniﬁcation of the image due to
lens no. 1 only. Express your answers accurately to four signiﬁcant ﬁgures. Put your ﬁnal answers in the spaces provided. Determine the locations of the parallel ray, the chief ray, and the focal ray for the
object and lens no. 1 only. Draw these three rays on the middle ﬁgure. Show the image due to lens no. 1 only. Calculate, showing all work, the location and magniﬁcation of the ﬁnal image due
to lens no. 1 and lens no. 2 acting together. Use the image from lens no. 1 as the object
for lens no. 2. Express your answers accurately to four signiﬁcant ﬁgures. Put your ﬁnal answers in the spaces provided. Determine the locations of the parallel ray, the chief ray, and the focal ray for lens
no. 2 and the ﬁnal image. Draw these three rays on the lower ﬁgure. Show the ﬁnal im age due to the combination of the lenses. Location of image due to lens no. 1 only is mm measured (left or right,
circle one) from lens no. 1. Magniﬁcation of image due to lens no. 1 only is Location of ﬁnal image due to lens no. 1 and lens no. 2 acting together
is mm measured (left or right, circle one) from lens no. 2. Magniﬁcation of image due to lens no. 1 and lens no. 2 acting together
is Thick Lens Imaging A Nikkor—H 85 mm f / 1.8 lens is to be used for making photomasks for integrated
circuit manufacture. The “optical formula” for this lens is shown below. All dimensions
given are in mm. The light used is to be of wavelength of 546.1 nm (e—line of mercury
lamp). The photomask artwork is placed to the left of the lens shown in the diagram below. A real image is formed at a reproduction ratio of 10 to 1. Calculate, showing all work, the distance of the object from the left—most glass
surface. Calculate, showing all work, the distance of the image from the rightmost glass
surface. Express your answers in mm accurately to within i0.1 mm. Put your answers in the spaces provided. Distance of object from left—most glass 2 mm Distance of image from right—most glass 2 mm Microscope Objective A microscope is constructed by combining a microscope objective and an eyepiece
in an appropriate way. A particular microscope objective examined in class has engraved
on it the following: Apo
10 / 0.25
160 / 0.17 Explain completely the meaning of each engraved marking by ﬁnishing the sentence started below. Add further complete sentences in each case as necessary. “Apo ” means “10” means “0.25 ” means “160” means “0.17” means Hemispherical Solid Immersion Lens  Magniﬁcation A hemispherical solid immersion lens for reading compact disks is shown in the
ﬁgure. The refractive index of the lens is m. The center of curvature of the lens lies on
the vertical ﬂat surface at the intersection of the dashed lines. The center of curvature is
taken as the origin of an (113,14) coordinate system (+11: to the right, +y up). The radius
of curvature is R. The refractive index of the surrounding medium is n2. An object on the compact disk is at the ﬂat surface of the lens extending from
the axis of the lens to its tip at height h. Two rays from the tip of this object are shown
in the diagram. In the region of refractive index 712, the equation for the upper ray is sin/61 R = _ t —— ——
581 CO (/61 0103/1 + 82.71061 _ a1) , where a1 and ,61 are deﬁned in the drawing. In the region of refractive index 112, the equation for the lower ray is $2 = ——cot/62 3/2 + R, where ,62 is deﬁned in the drawing. Calculate, showing all work, an analytic expression for the linear magniﬁcation m
as a function of n1, 712, R, and h only (eliminate all angles). Do not use the paraxial ray
approximation. Put your ﬁnal answer in the space provided below. 3
 Reﬂection from a Metal Linearly polarized light in air is reﬂected from the plane surface of a metal (a
highly lossy material). The incident linear polarization is at a 45° angle with repect to
the plane of incidence. Thus, there are equal amplitudes of TE and TM polarization in—
cident upon the metal. Of all the possible angles of incidence, there is one particular
angle that is designated as the “principal angle of incidence” and it is designated as 0?.
Corresponding to the prinicpal angle of incidence there is the “principal azimuthal an gle” and it is designated as «#1,. Give deﬁnitions of 0,, and 111?. Use mathematical expressions and complete sen tences as appropriate. Deﬁnition of 01,: Deﬁnition of app: General Optical Engineering Questions 1. A light ray is incident at an oblique angle on a boundary between dielectrics. The
incident light ray is propagating in a relatively low refractive index dielectric. On the
other side of the boundary is a relatively high refractive index dielectric. The refracted ray in the relatively high refractive index dielectric is (Circle one.)
a) deviated toward the normal to the boundary 2. The linear magniﬁcation (or transverse magniﬁcation or lateral magniﬁcation) of an
optical system is m, Where m = sl/s. The longitudinal magniﬁcation, on the other
hand, is As’/As. Write an expression for the longitudinal magniﬁcation as a function of m. The longitudinal magniﬁcation As’/As = — m2. 3. Of the aberrations coma and astigmatism, which is more important to be corrected in an astronomical telescope? (Circle one.)
a) coma 4. Linearly polarized light is propagating in glass of refractive index of 1.50. This light
is incident normally upon plane boundary with air (refractive index of 1.00). Immedi—
ately upon reflection, the electric ﬁeld of the electromagnetic wave has a phase shift (rel— ative to the phase of the incident wave) of (Circle one.)
. a) 0 degrees 5. A TE polarized light wave is propagating in a dielectric of refractive index of m. This light is incident upon plane boundary with a second dielectric of refractive index n2
(m > n2). The angle of incidence is greater than the critical angle. Immediately upon
reﬂection, the electric ﬁeld of the electromagnetic wave has a phase shift (relative to the phase of the incident wave) of (Circle one.)
c) 0 < phase shift < 180° 6. Circularly polarized light of freespace wavelength A can be converted to linearly po—
larized light by which of the following waveplates designed for that wavelength. (Circle one. )
b) quarter waveplate 7. For A being the freespace wavelength, in a typical glass of refractive index n in the visible portion of the electromagnetic spectrum (Circle one.)
a) dn/dA < 0 Imaging with Two Lenses f1=+100mm
f2=+50mm
Lens No. 1
51=+150mm
1+3. _ i
31 81' f1
51': flsl =‘300mm
81—f1
m =__8_1_’ = _ﬂ9 = _2
1 31 150
Lens No.2
32=—100mm
1 1 _ 1
32 82' f2
52’ = f2” = 33.333mm
82f2
_ 32' _ 33.333 _
m2— 82 _ _100 — +0.33333 Magniﬁcation of Lens No. 1 and Lens No. 2 acting together m = m1 m2 = —0.66667 Lens Lens
No. 1 No. 2 Thick Lens Imaging Real image, m < 0
Linear magniﬁcation, m = —0.10 = —1/10 Focal length, f = 85.0mm Object distance 1
s = (1 — Rdf = 11f = 11 (85.0mm) = 935mm
Distance of object from leftmost glass 2 935 — (85 — 54.2) = 904.2 mm Image distance
3’ 2 —m3 = —(—1/10) 935 mm = 93.5 mm Distance of image from rightmost glass 2 93.5 — (85 — 43.2) = 51.7mm Microscope Objective “Apo” means the objective is corrected for chromatic aberration at blue (486.1 nm), yel
low (587.6 nm), and red (656.3 nm) wavelengths. “10” means the linear magniﬁcation is 10X (m = 10) when the objective is used as in
tended. “0.25” means the numerical aperture is 0.25 (NA = 0.25). In air the objective accepts a
.cone of light with apex angle 2 sin‘1(0.25) = 28.96”. “160” means the tube length is 160 mm (L = so’ — f0). This is{ the extrafocal image
distance when the objective is used as intended. “0.17” means that the objective was designed to be used with a cover glass of thickness
0.17 mm; The use of a cover glass is needed in biological and medical applications. Hemispherical Solid Immersion Lens  Magniﬁcation Upper ray
:1: — cot(ﬂ a ) + Sin ﬁl R
1 1 1 m sin(ﬁ1 — a1) ’
. h  . .
szn a1 = E for angle of 1nc1dence a1
a — sin"1 1"—
1 _ R
n1 sin (11 = n2 sin ﬂl Snell’s law
_1 n ‘ ' _1 n h
= — s n = s n
ﬁl sm (“2 i a1) 2 (“2 R)
Lower ray
$2 = _COtﬁ2 212 + R)
h . .
tan a2 = E for angle of 1nc1dence a2
a — tan"1 2
2 ‘ R
n1 sin (12 = n2 sinﬁz Snell's law n1 n1 h __  —1  _  —1
'62 — S'ln S’L‘n a2) — S’I/n The magniﬁed image occurs at the intersection of the two ray lines. $1=$2=$m and y2:y1:ym sin ﬂl
sin(,61 — m) R cot(ﬁ1 — GIN/m + cot.32 31m + R and so
RU — ﬂ] 9'" 2 com“; — (:0th T a1) RU * 3%] m = Mhlcotﬁz — coma; 501)] where a1, [91, and 32 are all gii'en as functions of n1, 123, R, and h aboVe. Reﬂection from a Metal Deﬁnition of 01,: The principal angle of incidence, 0p, is the angle of incidence for which ¢TM—¢TE = 90°,
where ¢T M and ¢TE are the phase shifts upon reﬂection of the TE and TM polarized
components of the incident beam. Of all the angles of incidence, this angle produces the most nearly circularly polarized reﬂected light. The reﬂected elliptical polarization reaches its greatest width at this angle of incidence. Deﬁnition of 1,01,: The principal azimuthal angle is 11),, = tan—1ITTTTTA§IL where ’I‘T M and rTE are the am plitude reﬂection coefﬁcients when the angle of incidence is 01,. The principal azimuthal angle is thus a measure of the ellipticity, III—Mi, of the elliptical polarization. Considering
ITTEI . all possible angles of incidence, the principal azimuthal angle is the minimum value of the azimuthal angle. ...
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 Spring '08
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