Lecture 41: WED 29 APR Lecture 41: WED 29 APR

Physics 2102Jonathan Dowling

Ch. 36: DiffractionCh. 36: Diffraction

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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: =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?

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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 Maps2009

Review of Interference from Thin Films

Fig. 35-15

12 ?φ =

0 ≈

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Soap Bubble

Oil Slick

Reflection Phase Shifts

n1 n2

n1 > n2

n1 n2

n1 < n2

Reflection Reflection Phase ShiftOff lower index ---> 0

Off higher index ---> 210

(35-15)

High-To-Low: Phase Shift — NO!

Low-To-High: Phase Shift — Pi!

High-To-Low: Phase Shift — NO!

Low-To-High: Phase Shift — Pi!

02

210

Equations for Thin-Film Interference

Fig. 35-17

Three effects can contribute to the phase difference between r1 and r2.

1. Differences in reflection conditions.

2. Difference in path length traveled.

3. Differences in the media in which the waves travel. One must use the wavelength in each medium (/ n) to calculate the phase.

2

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

2 2 nL = × ×

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

2

0

22 integer wavelength = integer (out-of-phase waves)nL = × ×

22

n n

= ( )12

2

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

λ= + = K

2

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

= = K

(35-16)

Film Thickness Much Less Than

If L is much less than l, for example L < 0.1, then phase difference due to the path difference 2L can be neglected.

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

r2

r1

(35-17)

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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)/n2

Reflective 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 red light but half-integer wavelengths for blue.

Furthermore, the path difference 2L will change when light strikes the surface at different angles, again changing the interference condition for the different wavelengths of light.

(35-18)

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Problem Solving Tactic 1: Thin-Film Equations

Equations 35-36 and 35-37 are for the special case of a higher index film flanked by air on both sides. For multilayer systems, this is not always the case 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)