Lecture 29Lecture 29

Physics 2102Jonathan Dowling

Ch. 36: DiffractionCh. 36: Diffraction

Things You Should Learn fromThis Lecture

1. When light passes through a small slit, is spreads out and produces adiffraction pattern, showing a principal peak with subsidiary maximaand minima of decreasing intensity. The primary diffraction maximumis twice as wide as the secondary maxima.

2. We can use Huygens’ Principle to find the positions of the diffractionminima by subdividing the aperture, giving θmin = ±p λ/a, p = 1, 2, 3, ... .

3. Calculating the complete diffraction pattern takes more algebra, andgives Iθ=I0[sin(α)/α]2, where α = π a sin(θ)/λ.

4. To predict the interference pattern of a multi-slit system, we mustcombine interference and diffraction effects.

Single Slit Diffraction

When light goes through anarrow slit, it spreads outto form a diffractionpattern.

Analyzing Single Slit Diffraction

For an open slit of width a, subdivide the opening into segments andimagine a Hyugen wavelet originating from the center of each segment. Thewavelets going forward (θ=0) all travel the same distance to the screen andinterfere constructively to produce the central maximum.

Now consider the wavelets going at an angle such that λ = a sin θ ≅ a θ.The wavelet pair (1, 2) has a path length difference Δr12 = λ/2, andtherefore will cancel. The same is true of wavelet pairs (3,4), (5,6), etc.Moreover, if the aperture is divided into p sub-parts, this procedure can beapplied to each sub-part. This procedure locates all of the dark fringes.

thsin ; 1, 2, 3, (angle of the p dark fringe)p p

p pa

!" "= # = L

Conditions for DiffractionMinima

th

sin ; 1, 2, 3,

(angle of the p dark fringe)

p pp pa

!" "= # = L

Pairing and Interference Can the same technique be used to find the maxima, by choosing pairsof wavelets with path lengths that differ by λ? No. Pair-wise destructive interference works, but pair-wiseconstructive interference does not necessarily lead to maximumconstructive interference. Below is an example demonstrating this.

Calculating theDiffraction Pattern

We can represent the light through theaperture as a chain of phasors that “bends” and“curls” as the phase Δβ between adjacentphasors increases. β is the angle between thefirst and the last phasor.

Calculating theDiffraction Pattern (2)

( )2 sin / 2E r! "=

max max/ ; /E r r E! != =

( )max

max

sinsin / 2

/ 2

EE E!

"#

# "= =

sin2

a! "# $

%& =

2

max

sinI I!

"

"

# $= % &

' (

: or sinMinima m a m! " # $= ± = ±

2I CE=

Diffraction Patterns

sina!

" #$

=

2

max

sinI I!

"

"

# $= % &

' (

- 0.03 - 0.02 - 0.01 0 0.01 0.02 0.03

0.2

0.4

0.6

0.8

1

λ = 633 nm

a = 0.25 mm

0.5 mm

1 mm

2 mm

θ (radians)

The wider the slit opening a, or thesmaller the wavelength λ, the narrowerthe diffraction pattern.

Blowup

X-band: λ=100m

K-band:λ=10m

Ka-band:λ=1m

Laser:λ=1 micron 

Radar: The Smaller The Wavelength the Better The Targeting Resolution

0.005 0.01 0.015 0.02 0.025 0.03

0.01

0.02

0.03

0.04

0.05

Angles of the Secondary Maxima

The diffractionminima are precisely atthe angles wheresin θ = p λ/a and α = pπ(so that sin α=0).

However, thediffraction maxima arenot quite at the angleswhere sin θ = (p+½) λ/aand α = (p+½)π(so that |sin α|=1).

0.017350.0174150.014170.0142440.010990.0110830.007780.0079120.004530.004751θMax(p+½) λ/ap

1

23 4 5

λ = 633 nma = 0.2 mm

To find the maxima, one must look near sin θ = (p+½) λ/a, for placeswhere the slope of the diffraction pattern goes to zero, i.e., whered[(sin α/α)2]/dθ = 0. This is a transcendental equation that must be solvednumerically. The table gives the θMax solutions. Note that θMax < (p+½) λ/a.

θ (radians)

2

max

sinI I!

"

"

# $= % &

' (

Example: Diffraction of a laserthrough a slit

Light from a helium-neon laser (λ = 633 nm) passes through a narrowslit and is seen on a screen 2.0 m behind the slit. The first minimum ofthe diffraction pattern is observed to be located 1.2 cm from the centralmaximum. How wide is the slit?

11

(0.012 m)0.0060 rad

(2.00 m)

y

L! = = =

74

3

1 1

(6.33 10 m)1.06 10 m 0.106 mm

sin (6.00 10 rad)a

! !

" "

##

#

$= % = = $ =

$

1.2 cm

Width of a Single-SlitDiffraction Pattern

; 1, 2,3, (positions of dark fringes)p

p Ly p

a

!= = L

2(width of diffraction peak from min to min)

Lw

a

!=

w

-y1 y1 y2 y30

Exercise

Two single slit diffraction patterns are shown. The distancefrom the slit to the screen is the same in both cases. Which of the following could be true?

λ1

λ2

(a) The slit width a is the same for both; λ1>λ2.(b) The slit width a is the same for both; λ1<λ2.(c) The wavelength is the same for both; width a1<a2.(d) The slit width and wavelength is the same for both; p1<p2.(e) The slit width and wavelength is the same for both; p1>p2.

θ (degrees)

θ (degrees)

θ (degrees)

Combined Diffraction andInterference

So far, we have treateddiffraction and interferenceindependently. However, ina two-slit system bothphenomena should bepresent together.

( )( )

2

2

2slit 1slit

sin4 cos ;

sin ;

sin .

I I

a ay

L

d dy

L

!"

!

# #! $

% %# #

" $% %

& '= ( )

* +

= =

= =

d

a a

Notice that when d/a is an integer, diffraction minima will fall on top of“missing” interference maxima.

Interference Only

Diffraction Only

Both

Circular Apertures

When light passes through a circular aperture instead of a vertical slit,the diffraction pattern is modified by the 2D geometry. The minima occurat about 1.22λ/D instead of λ/a. Otherwise the behavior is the same,including the spread of the diffraction pattern with decreasing aperture.

Single slit of aperture aHole of diameter D

The Rayleigh Criterion

The Rayleigh ResolutionCriterion says that the minimumseparation to separate two objectsis to have the diffraction peak ofone at the diffraction minimum ofthe other, i.e., Δθ = 1.22 λ/D.

Example: The Hubble Space Telescopehas a mirror diameter of 4 m, leading toexcellent resolution of close-lyingobjects. For light with wavelength of500 nm, the angular resolution of theHubble is Δθ = 1.53 x 10-7 radians.

Example

A spy satellite in a 200km low-Earth orbit is imaging theEarth in the visible wavelength of 500nm.

How big a diameter telescope does it need to read anewspaper over your shoulder from Outer Space?

Δθ = 1.22 λ/D

Letters on a newspaper are about Δx = 10mm apart.Orbit altitude R = 200km & D is telescope diameter.

Christine’s Favorite Formula:Δx = RΔθ = R(1.22λ/D)

D = R(1.22λ/Δx)

= (200x103m)(1.22x500x10–9m)/(10X10–3m)

= 12.2m

Example Solution

R

Δx

Δθ

Los Angeles from Space!