Lecture 27: THU 29 APR 2010Lecture 27: THU 29 APR 2010

Physics 2102Jonathan Dowling

Ch. 36: DiffractionCh. 36: Diffraction

QuickTime™ and a decompressor

are needed to see this picture.

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

⎛ ⎞= ⎜ ⎟⎝ ⎠

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

X-band: =10cm

You are doing 137 mph on I-10 and you pass a little old lady doing 55mph when 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?

QuickTime™ and a decompressor

are needed to see this picture.

QuickTime™ and a decompressor

are needed to see this picture. 1m

1m

10 m1000m

No! Beam is much wider than car separation —

too wide to tell who is who.

No! Beam is much wider than car separation —

too wide to tell who is who.

w =2La

=2 ×0.000001m×1000m

0.1m=0.02m

Laser-band: =1m

Yes! Beam width is much less

than car separation.

Yes! Beam width is much less

than car separation.

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?

1

2

(a) The slit width a is the same for both; 12.(b) The slit width a is the same for both; 12.(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

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

Diffraction Gratings

Fig. 36-17 Fig. 36-18

sin for 0,1, 2 (maxima-lines)d m m = = K

(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 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., 122D.

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?

122D (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.

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

Some Professor’sHouse: Google Maps