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