Lecture 41: WED 29 APRLecture 41: WED 29 APR

Physics 2102Jonathan Dowling

Ch. 36: DiffractionCh. 36: Diffraction

PHYS 2102-2 FINAL

• 5:30-7:30PM FRI 08 MAY• COATES 143• 1/2 ON NEW MATERIAL• 1/2 ON OLD MATERIAL• Old Formula Sheet:

http://www.phys.lsu.edu/classes/spring2007/phys2102/formulasheet.pdf

X-band: λ=10cm

K-band:λ=2cm

Ka-band:λ=1cm

Laser:λ=1 µm 

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

!= = !

2(width of diffraction peak from min to min)

Lw

a

!=

w

-y1 y1 y2 y30

X-band: λ=10cm

You are doing 137 mph on I-10 and you pass a little old lady doing 55mphwhen a cop, Located 1km away fires his radar gun, which has a 10 cm opening.Can he tell you from the L.O.L. if the gun Is X-band? What about Laser?

1m

1m

10 m1000m

w =2!L

a=2 " 0.1m "1000m

0.1m= 2000m

No! Beam is much wider than carseparation —

too wide to tell who is who.

w =2!L

a=2 " 0.000001m "1000m

0.1m= 0.02m

Laser-band: λ=1µm

Yes! Beam width is much less than

car separation.

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

A device with N slits (rulings) can be used to manipulate light, such as separatedifferent wavelengths of light that are contained in a single beam. How does adiffraction grating affect monochromatic light?

Diffraction Gratings

Fig. 36-17 Fig. 36-18

sin for 0,1, 2 (maxima-lines)d m m! "= = …

(36-11)

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 (The smaller the wavelength or the bigger thetelescope opening — the better the angular resolution.)

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

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!Corona Declassified Spy Photo:Circa 1960’s

SomeProfessor’sHouse: GoogleMaps2009

Review of Interference from Thin Films

Fig. 35-15

12?! =

0! "

Soap Bubble

Oil Slick

Reflection Phase Shifts

n1 n2

n1 > n2

n1 n2

n1 < n2

Reflection Reflection Phase ShiftOff lower index ---> 0Off higher index ---> λ/2 = 180° = π

(35-15)

High-To-Low: Phase Shift — NO!

Low-To-High: Phase Shift — Pi!

λ = 360° = 2π

λ/2 = 180° = π

Equations for Thin-Film Interference

Fig. 35-17

Three effects can contribute to the phasedifference between r1 and r2.

1. Differences in reflection conditions.

2. Difference in path length traveled.

3. Differences in the media in which the wavestravel. One must use the wavelength in eachmedium (λ / n) to calculate the phase.

2

odd number odd number2 wavelength = (in-phase waves)

2 2n

L != " "

½ wavelength phase difference to difference in reflection of r1 and r2

2

! 0

22 integer wavelength = integer (out-of-phase waves)n

L != " "

2

2

n

n

!! = ( )1

2

2

2 for 0,1,2, (maxima-- bright film in air)L m mn

!= + = …

2

2 for 0,1,2, (minima-- dark film in air)L m mn

!= = …

(35-16)

Film Thickness Much Less Than λ

If L is much less than l, for example L < 0.1λ, then phasedifference due to the path difference 2L can beneglected.

Phase difference between r1 and r2 will always be ½wavelength → destructive interference → film will appeardark when viewed from illuminated side.

r2r1

(35-17)

Top of Soap BubbleFilm is So ThinIt Looks DarkAt All Wavelengths:Anti-Reflective Coating!

As You Go DownThickness Increases Steadily (Gravity)And DifferentWavelengths hitReflection Condition2L=(m+1/2)λ/n2Reflective Coating

Color Shifting by Morpho Butterflies and Paper Currencies

Fig. 35-19

For the same path difference, different wavelengths(colors) of light will interfere differently. For example,2L could be an integer number of wavelengths for redlight but half-integer wavelengths for blue.

Furthermore, the path difference 2L will change whenlight strikes the surface at different angles, againchanging the interference condition for the differentwavelengths of light.

(35-18)

Problem Solving Tactic 1: Thin-Film Equations

Equations 35-36 and 35-37 are for the special case of a higher index filmflanked by air on both sides. For multilayer systems, this is not always thecase and so these equations are not appropriate.

What happens to these equations for the following system?

n1=1 n2=1.5 n3=1.7

r1

r2

L

(35-19)