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Unformatted text preview: ECE 414 Exam 1
2.24.11 Name: 60 {Vii/29474 Signature: 92' 99' 9'0 H You are allowed to use one 8.5” x 11” sheet of paper, a (non—programmable)
calculator, pen / pencil to complete this exam. Be sure to Show all of your work. Answers without work will receive no credit. 1. (15 Points) A lens is used to form an image ofa tree as shown in the diagram below.
Using this image, answer the following questions. a) On the diagram above, draw the three principle rays to locate where the image of
the tree will be formed. Be sure to label any critical points required to draw the
principle rays. b) Calculate the exact distance between the lens and the image plane analytically. 1.1,.—LVJ~._L .1
35+ ’P 75"1" ‘3:— 4g"( ' 5
O ‘ 4‘ $0
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c) Calculate the transverse magnification of the image. ’ (O n
M _ " {1. " ‘
T 50 Ol‘l'IO ‘ mm
d) Is the image upright or inverted? am/ wt 5044 9% {4‘ .3 ,‘ﬁmfw/ #mm pr (at) 2. (30 Points) — Optical systems with many optical components can be greatly simplified by
representing each optical element as a matrix. These matrices can be combined into
a single "Transmission Matrix". Answer the following questions using the optical
system shown below. § W
L22m L=5m a) In the proper order, write out the individual matrices needed to calculate the
transmission matrix for the system between dashed lines. /n reverse mew: R=2m 1 1 l O l S l O l .1 l
.L m a
0  0 {.3 0 l 0 O ( Lj'll b) What is the total transmission matrix for this optical system? r015 *0345 c) Using this transmission matrix, write an equation relating the ray output to the ray Input 20W = Qm + 7‘ 1? RM! Em11L l ll 7 R M p Ra”: “0.25 “(HHS a ' Rout, : 11°75“ w 0‘7‘75 9"",
m d) A ray incident on this optical system has a slope of 0.2, and height of 0.02m from
the center of this system. Calculate the output slope and height. QM: may Mam: L456," Raﬁ M25  Ma 0.?45 m: '0. {6‘1 ’x.yﬁﬁ,mw . . I, _ r. ..._v .. «wwnwn—«Wrﬁ
y
., 3. . (15 Points] i Answer the following questions regarding a plane wave in the following form:
(Be sure to include units in answers) ' a,
% g=zgoasw62) =2 w: L/Yzo'sir , />’= Mole a) E = Sﬁcos[4 x10157r(t— 2'x10*8z)] _ 7 :3 With E in volts/meter, t in second, 2 in meters ""7 a) Fill in the blanks: a.» X i) The electric field is polarized in the a direction.
ii) The magnetic field is polarized in the 5? direction.
iii) The field is propagating in the 1 direction. b) What is the frequency of this field's oscillation? a)? Q‘ﬂg ~‘2 15‘: y : “Hawk: ‘5
‘ a? 2? 2*!0 HZ c) What is the wavelength? A,
k: 3‘3 : M s '5 6 ? mi? 91:10} 5m" d) What is the propagation velocity of the wave? 7?; 4/.» ?  ‘1 = W
T: ﬁlm“? : 14 = 35x/0‘92H0'5= 5‘ [ow/5 e) What is the refractive index of the material through which this wave is travelling? Ag:
“w W»?
,. '3‘108’” \
' E755?“ 1:14 f) What is the intensity of this wave? , 1‘ _l_ _ , '1
1: +£6.66) ~ 1 ‘1735‘1Xl0'13[03 gamma; Wm g) Sketch one period of this electric field labeling axes, amplitude, and wavelength. *5
s l
l
l
I 4. [10 Points] Shown below is a MachZender interferometer. The laser source has a wavelength of
532nm, and coherence length of 3cm. Notice that one “arm” of the interferometer
has an unknown refractive index, while the other arm contains air. Mirror a) What is the optical path length for each arm of the interferometer? 2‘5cm: l00m
0M2” 2‘5'n3 l0n 6m b) What is the largest refractive index the material could have in order to still
observe interference in this system? ‘ ln arm Z3 05% Wm/ AOPL4 LL=3M mm: (om; tom= 3am
zoom): 3 nl = 05
h=L3 Whlbmr n 11.3 \v ,5. ‘ [30 Points) A Gaussian beam. with lambda = 543nm passes from air into a thin glass lens with
refractive index n = 1.5, and focal distance f = 5mm, then back into air; At the left
side of the lens, the beam has a radius of .2cm, and 40cm radius of curvature. Find the beam radius and the radius of curvature of the Gaussian beam emerging from this thin lens. W lemg all WM fax/96 i0 male/5
W i‘ 0 ~ '1 0 , : _, €L.“ ‘ (“D Mos l ‘100 l
{Balms “L: L i : J,“ 543*(0'4
{an 2 two WI Tami):
 l J. : r ‘043 ~>7 ~: ———~ gnn Q: 5 J a 9\ glh 2‘51; O‘OQEDT . = Xx" ‘ 43w Law _ gm‘ “3%? 40006“ g \, #:mﬂ‘a _ : 03‘7W“J0*0°“ =~0.005l+ail.l1ﬂ0'b
w+ 5% * D maﬁa“ '900(0.3W3+J'o.ooéql)+l / .L ., . 3  )\
gwlr" 5 E ’} ﬁnk) l Q_A 0:043; 2 ‘T'PLWl ‘4 (U ‘ S
‘  : 00 :2
‘U “ V‘MM‘BQ) 0‘ M 3980410 3% ‘Hnlg 33 a “ern [mg’m OlaMl Qiﬂgwl’ baa/narmdu‘ugc‘ﬂi ...
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 Spring '11
 Alenxendra

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