Lecture 40: MON 27 APR Lecture 40: MON 27 APR

Physics 2102Jonathan Dowling

Ch. 36: DiffractionCh. 36: Diffraction

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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

⎛ ⎞= ⎜ ⎟⎝ ⎠