221 lecture 8 s12.ppt - Texas A&M...
Transcript of 221 lecture 8 s12.ppt - Texas A&M...
DiffractionDiffraction
February 9, 2012Chapter 36Chapter 36
February 9, 2012 Physics 221 1
Diffraction and coherent sources
Reminder of the form of E&M WavesE(x t) = E0 j sin(kx - ωt + ϕ) withE(x,t) E0 j sin(kx ωt + ϕ) , with k = 2π/λ and ω = 2πf
February 9, 2012 Physics 221 2
Finding the resultant E field
Ep is the resultant field
February 9, 2012 Physics 221 3
Intensity pattern for a single slit
]/)(sinsin[2
⎬⎫
⎨⎧
=λθπaII
n diffractioslit singlein intensity
/)(sin0
⎭⎬
⎩⎨=
λθπaII
gy
sinθa
fringesdarktheofpositionfor the
m sin=
λθa
February 9, 2012 Physics 221 4
fringesdark theofposition for the
Photo of Single Slit Pattern
λ
fringedark1stofpositionfor the
1 aλθ =
February 9, 2012 Physics 221 5
fringedark 1st ofposition for the
Diffraction patterns for 2 slits
βφ )2/sin(cos2
20II ⎥
⎤⎢⎡
⎟⎞
⎜⎛=
βwidthfiniteofslitsfor two2/2
cos0II ⎥⎦
⎢⎣
⎟⎠
⎜⎝
θπβθπφ sin2andsin2 widthfiniteofslitsfor two
ad== θ
λβθ
λφ sinand sin
February 9, 2012 Physics 221 6
Now let’s look at what happens with several slits..
ceinterferen veconstructifor sin λθ md =
cee e eveco s uco
Slits are narrower than λso the diffraction patterns arepbroad and modulated by the interference effects from multiple slits.
February 9, 2012 Physics 221 7
Diffraction patterns for different numbers of slits
February 9, 2012 Physics 221 8
Diffraction gratings
sin λθ md =
anglesat be willbandsBright ceinterferenveconstructifor
condition. thissatisfying
February 9, 2012 Physics 221 9
Grating Spectrograph Can change the orientationof the beam to the gratingto map out the spectrum ofthe sample beam of light.
February 9, 2012 Physics 221 10
X-Ray Diffraction
Crystals are a three dimensional diffraction grating.
February 9, 2012 Physics 221 11
The “Bragg Condition” 2dsinθ = mλ gives constructive interference
February 9, 2012 Physics 221 12
Circular apertures and resolving power
22.1 sin 1 dλθ =
aperature.circularafrom ringdark first theof angle
aperature.circular a from
February 9, 2012 Physics 221 13
Circular apertures and resolving power an example
Problem 36.48: A converging lens 7.20 cm in diameter has a focal length of 300 mm. If resolutionis diffraction limited, how far away can an object beif points on it 4 00 mm apart are to be resolved
m9550 ⎞⎛ Eλ
if points on it 4.00 mm apart are to be resolved.Use λ = 550 nm.
li itdiff tit thi ttb t thl
632.9m220.7m955022.122.1 sin 1 −=⎟
⎠⎞
⎜⎝⎛
−−
== EEE
dλθ
m429 ;632.9bjdi300.4
limit.n diffractioat thepointstwoebetween th angle
=−=− dEE
February 9, 2012 Physics 221 14
;object todistance