Lecture 26: TUE 27 APR 2010 Lecture 26: TUE 27 APR 2010

Physics 2102Jonathan Dowling

Ch. 36: DiffractionCh. 36: Diffraction

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EXAM III

• AVG = 60• STD = 15• A=100–85• B=84–75• C=74–45• D=44–40• F=39–0

Michelson Interferometers:

As we saw in the previous example, interference is a spectacular wayof measuring small distances (like the thickness of a soap bubble), sincewe are able to resolve distances of the order of the wavelength of thelight (for instance, for yellow light, we are talking about 0.5 of a millionth of a meter, 500nm). This has therefore technological applications.

In the Michelson interferometer, light from asource (at the left, in the picture) hits a semi-plated mirror. Half of it goes through to the right and half goes upwards. The two halves are bounced back towards the half plated mirror, interfere, and the interference can be seen by the observer at the bottom. The observer will see light if the two distances travelled d1 and d2 are equal, and will see darkness if they differ by half a wavelength.

Michelson-Morley ExperimentMichelson won the Nobel prize in 1907, "for his optical precision instruments and the spectroscopic and metrological investigations carried out with their aid"

"The interpretation of these results is that there is no displacement of the interference bands. ... The result of the hypothesis of a stationary ether is thus shown to be incorrect." (A. A. Michelson, Am. J. Sci, 122, 120 (1881))

The largest Michelson interferometer in the world is in Livingston, LA,in LSU owned land (it is operated by a project funded by the NationalScience Foundation run by Caltech and MIT, and LSU collaborates in the project).

http://www.ligo-la.caltech.edu

Mirrors are suspended with wires and will movedetecting ripples inthe gravitational field due to astronomical events.

Gravitational Waves Interferometry:an International Dream

GEO600 (British-German)Hannover, Germany

LIGO (USA)Hanford, WA and Livingston, LA

TAMA (Japan)Mitaka

VIRGO (French-Italian)Cascina, Italy

AIGO (Australia), Wallingup Plain, 85km north of Perth

Things You Should Learn from This Lecture

1. When light passes through a small slit, is spreads out and

produces a diffraction pattern, showing a principal peak with

subsidiary maxima and minima of decreasing intensity. The

primary diffraction maximum is twice as wide as the secondary

maxima.

2. We can use Huygens’ Principle to find the positions of the

diffraction minima by subdividing the aperture, giving min= ±p

/a, p = 1, 2, 3, ... .

3. Calculating the complete diffraction pattern takes more algebra,

and gives I=I0[sin()/]2, where = a sin()/.

4. To predict the interference pattern of a multi-slit system, we must

combine interference and diffraction effects.

Single Slit Diffraction

When light goes through a narrow slit, it spreads out to form a diffraction pattern.

Analyzing Single Slit Diffraction

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

Now consider the wavelets going at an angle such that a sin a . The wavelet pair (1, 2) has a path length difference r12 = , and therefore 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 be applied to each sub-part. This procedure locates all of the dark fringes.

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

= ≅ = L

Conditions for Diffraction Minima

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 pairs of wavelets with path lengths that differ by ? No. Pair-wise destructive interference works, but pair-wise constructive interference does not necessarily lead to maximum constructive interference. Below is an example demonstrating this.

Calculating theDiffraction Pattern

We can represent the light through the aperture as a chain of phasors that “bends” and “curls” as the phase between adjacent phasors increases. is the angle between the first and the last phasor.

Calculating theDiffraction Pattern (2)

( )2 sin / 2E rθ β=

max max/ ; /E r r E = =

( )maxmax

sinsin / 2

/ 2

EE Eθ

αβ

β α= =

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

= 633 nm

a = 0.25 mm

0.5 mm

1 mm

2 mm

(radians)

The wider the slit opening a, or

the smaller the wavelength , the

narrower the diffraction pattern.

Blowup

X-band: =10cm

K-band: =2cm

Ka-band:=1cm

Laser:=1 m 

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 diffraction minima are precisely at the angles wheresin = p /a and = p(so that sin =0).

However, the diffraction maxima are not quite at the angles where sin = (p+½) /aand = (p+½)(so that |sin |=1).

p (p+½) /a Max

1 0.004750.0045

3

2 0.007910.0077

8

3 0.011080.0109

9

4 0.014240.0141

7

5 0.017410.0173

5

1

23 4 5

= 633 nma = 0.2 mm

To find the maxima, one must look near sin = (p+½) /a, for places where the slope of the diffraction pattern goes to zero, i.e., whered[(sin )2]/d= 0. This is a transcendental equation that must be solved numerically. The table gives the Max solutions. Note that Max (p+½)/a.

(radians)

2

max

sinI I

⎛ ⎞= ⎜ ⎟⎝ ⎠