Lecture 29Lecture 29

Physics 2102Jonathan Dowling

Ch. 36: DiffractionCh. 36: Diffraction

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: =100m

K-band: =10m

Ka-band:=1m

Laser:=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 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

⎛ ⎞= ⎜ ⎟⎝ ⎠

Example: Diffraction of a laser through a slit

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

11

(0.012 m)0.0060 rad

(2.00 m)

y

L = = =

74

31 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 distance from the slit to the screen is the same in both cases. Which of the following could be true?

(a) The slit width a is the same for both; .(b) The slit width a is the same for both; .(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.

(degrees)

(degrees)

(degrees)

Combined Diffraction and Interference

So far, we have treated diffraction and interference independently. However, in a two-slit system both phenomena should be present together.

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 occur at about 1.22/D instead of /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 Resolution Criterion says that the minimum separation to separate two objects is to have the diffraction peak of one at the diffraction minimum of the other, i.e., D.

Example: The Hubble Space Telescope has a mirror diameter of 4 m, leading to excellent resolution of close-lying objects. For light with wavelength of 500 nm, the angular resolution of the Hubble is = 1.53 x 10-7 radians.

Example

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

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

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

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

Christine’s Favorite Formula:x = R = R(1.22/D)

D = R(1.22/x)

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

= 12.2m

Example Solution

R

x

Los Angeles from Space!Corona Declassified Spy Photo:Circa 1960’s