# Interference and Diffraction of EM waves

## Pełen tekst

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v

t E B

 

t J D

H

 

v v s

s

s L

s L

t dS J D

dl H

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v

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### The same phases The oposite phases

Constructive interference Destructive interference

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### The key principle: Huygen’s Principle

Christian Huygens 1629-1695

All points in a wavefront serve as point sources of spherical secondary waves.

After a time t, the new wavefront will be the tangent to all the resulting spherical waves.

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### Huygen’s Principle

For plane waves entering a single slit, the waves emerging from the slit start spreading out, diffracting

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### Young’s Double Slit Experiment

For waves entering two slits, the emerging waves interfere and form an interference (diffraction) pattern

Young experiment in 1801: light is wave phenomenon

First plane wave through a small slit yields coherent spherical wave

Then interposed two slits:

interference of two spherical waves on a screen

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### Interference

• Phase difference between two waves can change for paths of different lengths

• Each point on the screen is determined by the path length difference DL of the rays reaching that point

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12

1

1

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0

### When the interference is possible

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Red laser light (=633nm) goes through two slits d=1cm apart, and produces a diffraction pattern on a screen L = 55cm away. How far apart are the fringes near the center?

For the spacing to be 1mm, we need d~ L/1mm=0.35mm If the fringes are near the center, we can use

sin  ~ , and then

m=dsin~d => =m/d is the angle for each maximum (in radians)

D= /d =is the “angular separation”.

The distance between the fringes is then Dx=LD=L/d=55cm 633nm/1cm=35 mm

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### Example 2

In a double slit experiment, we can measure the wavelength of the light if we know the distances between the slits and the angular separation of the fringes.

If the separation between the slits is 0.5mm and the first order maximum of the interference pattern is at an angle of 0.059o from the center of the pattern, what is the wavelength and color of the light used?

d sin=m =>

=0.5mm sin(0.059o)

= 5.15 x 10-7m=515nm ~ green

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1 0

2 0

1

2 2 1

0 2

1 1 1

2 2 2

1 2

avg

0

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### Example

A double slit experiment has a screen 120cm away from the slits, which are 0.25cm apart. The slits are illuminated with coherent 600nm light. At what distance above the central maximum is the average intensity on the screen 75% of the maximum?

I/I0=4cos2f/2; I/Imax=cos2f/2 =0.75 => f=2cos–1 (0.75)1/2=60o=/3 rad f=(2d/)sin => = sin-1(/2d)f0.0022o40 mrad (small!)

y=L48mm

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## Diffraction

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### Huygen’s Principle

Christian Huygens 1629-1695

All points in a wavefront serve as point sources of spherical secondary waves.

After a time t, the new wavefront will be the tangent to all the resulting spherical waves.

(26)

### Huygen’s Principle

For plane waves entering a single slit, the waves emerging from the slit start spreading out, diffracting

(27)

### Young’s Double Slit Experiment

For waves entering two slits, the emerging waves interfere and form an interference (diffraction) pattern

Young experiment in 1801: light is wave phenomenon

First plane wave through a small slit yields coherent spherical wave

Then interposed two slits:

interference of two spherical waves on a screen

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1

2

1

2

1

2

### Locating the Minima

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Repeat previous analysis for pairs of rays, each separated by a vertical distance of a/2 at the slit.

Setting path length difference to /2 for each pair of rays, we obtain the first dark fringes at:

(first minimum)

### a     a   

For second minimum, divide slit into 4 zones of equal widths a/4 (separation between pairs of rays). Destructive interference occurs when the path length difference for each pair is /2.

(second minimum)

### a     a   

Dividing the slit into increasingly larger even numbers of zones, we can find higher order minima:

(minima-dark fringes)

### a   m  m 

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Diffraction pattern from a single narrow slit.

### Diffraction and the Wave Theory of Light

Central maximum

Side or secondary maxima

Light

Fresnel Bright Spot.

Bright spot

Light

These patterns cannot be explained using geometrical optics!

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### Single Slit Diffraction

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

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Here we will show that the intensity at the screen due to a single slit is:

m

2

###    

In Eq. 1 , minima occur when:

###  

If we put this into Eq. 2 we find:

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### Diffraction of a laser through a slit(example)

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?

1 1

7

4 3

1 1

1.2 cm

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p

###  

w

-y1 0 y1 y2 y3

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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?

1m 1m

1000m 10 m

w  2

### 

L

a 2  0.1m1000m

0.1m  2000m w 2aL 2 0.000001m0.1m 1000m 0.02m Laser-band: =1mm

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### Angles of the Secondary Maxima

The diffraction minima are precisely at the angles where

sin q = p l/a and a = pp (so that sin a=0).

However, the

diffraction maxima are not quite at the angles where sin q = (p+½) l/a

and a = (p+½)p (so that |sin a|=1).

p (p+½) /a Max 1 0.00475 0.00453 2 0.00791 0.00778 3 0.01108 0.01099 4 0.01424 0.01417 5 0.01741 0.01735

1

2

3 4 5

l = 633 nm a = 0.2 mm

2 max

I I sin

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### 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 a Hole of diameter D

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### 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  1.22

/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 D = 1.53 x 10-7 radians.

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3

–9

–3

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### Conventional photography

Light

Object

Reflected wave

Photographic film:

The intensity is recorded

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### • Produces interference pattern which is recorded,

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Reference wave

Photographic film.

Interference of reference and reflected waves is recorded

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### Types of holograms

Transmission hologram:

reference and object waves traverse the film from the same side

Reflection hologram:

reference and object waves traverse the emulsion from opposite sides

Invented by Benton, Can be reconstructed in normal light

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### diffraction minima of a single slit by subdividing the aperture

(minima-dark fringes)

2 1

0 2

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12

2 1

0 2

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m

2

m

2

2

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