6293 - Troubleshooting and Supporting Windows 7 in the Enterprise

6293 - Troubleshooting and Supporting Windows 7 in the Enterprise

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1800 U LEARN (1800 853 276) 6293 – Troubleshooting and Supporting Windows 7 in the Enterprise Vendor Course Code: 6293 Course Length: 3 days Overview: This course is designed for Information Technology (IT) professionals who have experience with Windows XP and Windows Vista who work as Windows 7 Enterprise Desktop Support Technicians (EDSTs) in Tier 2 support environments. The goal of this training is to enable these individuals to support the Windows 7 operating system and solve technical troubleshooting problems in a Windows 7 and Windows Server 2008 R2 networking environment.
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Nombre de lectures 29
Langue English
Poids de l'ouvrage 1 Mo
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`
Physics 202, Lecture 29
Today s Topics

  About the final exam
  Diffraction
  Single-slit Diffraction
  Reminder of Two-slit Interference
  Double-Slit Diffraction
  Diffraction on Circular Apertures
  The Rayleigh Criterion
  Resolving wavelengths z
About the Final Exam: Logistics
th  The exam will be at 7:25-9:25pm on Tuesday, Dec 20
  Rooms TBA by email
  Distribution of tests starts at 7:15pm.
  Four (3+1) 8½ x 11 single sided sheets are allowed.
  Put down whatever you like, prepare it yourself. (no
photocopying, download-printing of lecture notes/exam solutions/
examples. etc. )
  Any calculator is fine.
  A 2B pencil for Scantron.
  Contact us if special arrangements are necessary
  Other exams at the same time or conflicting university activities
  The Exam is cumulative.
  1-2 questions each for chapters 21-30
  3-4 questions each for 31-33
  Will send out practice questions on 31-33 by Friday
  There will be 25 multiple-choice problems. Where Are the Dark Fringes?
D
2
! $D sin(! /2)
I =I # &0
! /2" %
D
2"
! ' Dsin$
#D
D
Angular Separation
Between minima
Small angles: 2! ! The dark fringes occur at :
,
I=0 à sin(β/2)=0 à(π/λ)Dsinθ = mπ D Ddark
sinθ =mλ/D, m=± 1, ± 2, ± 3,...dark
 Central bright dot width Δθ =2 λ/D,First dark fringes at =± λ/D `
`
`
`
`
Lights As Rays?
 Due to diffraction, light beam of finite size does not travel
as perfect straight rays!
θ∼λ/a
a ~ a+2θL a
L
screen
 Numerical example: Estimate the size of laser beam on
screen (λ~600nm, L=1m)
-5a= 1cm, θ~ λ/a= 6x10 , a ~1.01cm à +1%
-4a= 2mm, θ~ λ/a= 3x10 , a ~2mm+0.6mm à +30%
-4a= 1mm, θ~ λ/a= 6x10 , a ~ 1mm+1mm à +100%
-3a= 0.1mm, θ~ λ/a= 6x10 , a ~0.1mm+12mm ! à +12000% Reminder: Two-slit Interference
path length difference
δ =dsinθ ~ dθ ~ d y/l
% ( Angular Separation "dsin#2" I = I cos ' * oBetween minima d & ) $
Small angles
!
! Two-slit Diffraction and Iterference
2# & sin(" /2)
I =I % ( 0 " /2$ '
2*
" ) Dsin,
+
# & "2I = I cos % ( o
$ ' 2
! 2)
" = dsin+
*
2# & * - sin("/2) )2I = I cos , / % ( ! 0 + . "/2 2$ '
! Diffraction on Circular Apertures

 Light through apertures will produce diffractive patters
depending on their shape. For circular apertures the
diffractive patters is made of concentric rings
Integral performed in 2D. λ
1.22 Gives slightly different First dark ring at θ=
D answer than for a slit. `
Resolution of Single-slit and Circular Apparatus
each smeared two Rayleigh s Criterion
due to diffraction separate beams
Minimally Not Separable
separable separable
 Single slit: θ = λ/D min
 Circular Aperture: θ = 1.22 λ/D minz
Example: Watching (buying!) an HDTV
  Human eye has a typical pupil diameter of about 5 mm.
  What is the minimum distance between two red (λ=700nm)
dots human eye can separate at 3 meters?
λ 700nm −4θ =1.22 =1.22 ≈1.7×10 Radmin
D 5mm
−4ΔS ≈θ × L ≈1.7×10 × 3000 mm =0.51 mmmin min

  Compare the above resolution to the pixel spacing of a 32
HDTV (720p or 1080p).

HDTVGeometricParameters
PixelSpacing(mm)
ScreenSize "720p" "1080p"
Width(mm) Height(mm)
(inch) (1280x720) (1920x1080)
32 707.1 398.3 0.55 0.37
47 1038.6 585.0 0.81 0.54Resolving Wave Lengths N slits
Maxima at y = mLλ/d

For two wavelengths of light.
Δy = mLλ /d-mLλ /d = mLΔλ/d 1 2

Want the second wavelength of light to show up one
small maxima from the peak. Δy ~ Lλ/Nd

mLΔλ/d = Lλ/Nd

Gives resolving power: R = λ/Δλ = mN