Interference and Diffraction. Certain phenomena require the light (the electromagnetic radiation) to...

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Interference and Diffraction

Certain phenomena require the light (the electromagnetic radiation)

to be treated as waves.

Two relevant examples are: interference and diffraction

Plane Electromagnetic Waves

Waves are in phase,but fields oriented at 900

Speed of wave is c

At all times E = c B

c m s 1 3 100 08/ /

E(x, t) = EP sin (kx-t)

B(x, t) = BP sin (kx-t) z

j

x

Ey

Bz

c

Plane Electromagnetic Waves

E and B are perpendicularto each other, and to the direction of propagationof the wave.

The direction of propagationis given by the right hand rule: Curl the fingers from Eto B, then the thumb pointsin the direction of propagation.

Electromagnetic wavespropagate in vacuumwith speed c, the speed of light.

Combination of Waves

In general, when we combine two waves to form a composite wave,the composite wave is the algebraic sum of the two original waves,point by point in space [Superposition Principle].

When we add the two waves we need to take into account their:

• Direction• Amplitude• Phase

+ =

Combination of Waves

The combining of two waves to form a composite wave is called:Interference

The interference is constructive if the waves reinforce each other.

+ =

Constructive interference(Waves almost in phase)

Combination of Waves

The combining of two waves to form a composite wave is called:Interference

The interference is destructive if the waves tend to cancel each other.

+ =

(Close to out of phase)

(Waves almost cancel.)

Destructive interference

Interference of Waves

+ =

Constructive interference(In phase)

+ =

( out of phase)

(Waves cancel)

Destructive interference

Interference of Waves

When light waves travel different paths, and are then recombined, they interfere.

Each wave has an electric field whose amplitude goes like:

E(s,t) = E0 sin(ks-t) î

Here s measures the distance traveled along each wave’s path.

Mirror

1

2*

+ =

Constructive interference results when light paths differ by an integer multiple of the wavelength: s = m

Interference of Waves

When light waves travel different paths, and are then recombined, they interfere.

Each wave has an electric field whose amplitude goes like:

E(s,t) = E0 sin(ks-t) î

Here s measures the distance traveled along each wave’s path.

Mirror

1

2*

Destructive interference results when light paths differ by an odd multiple of a half wavelength: s = (2m+1) /2

+ =

Interference of Waves

Coherence: Most light will only have interference for small optical path differences (a few wavelengths), because the phase is not well defined over a long distance. That’s because most light comes in many short bursts strung together.

Incoherent light: (light bulb)

random phase “jumps”

Interference of Waves

Coherence: Most light will only have interference for small optical path differences (a few wavelengths), because the phase is not well defined over a long distance. That’s because most light comes in many short bursts strung together.

Incoherent light: (light bulb)

Laser light is an exception: Coherent Light: (laser)

random phase “jumps”

Thin Film Interference

We have all seen the effect of colored reflections from thin oil films, or from soap bubbles.

Film; e.g. oil on water

Thin Film Interference

We have all seen the effect of colored reflections from thin oil films, or from soap bubbles.

Film; e.g. oil on water

Rays reflected off the lowersurface travel a longeroptical path than raysreflected off upper surface.

Thin Film Interference

We have all seen the effect of colored reflections from thin oil films, or from soap bubbles.

Film; e.g. oil on water

Rays reflected off the lowersurface travel a longeroptical path than raysreflected off upper surface.

If the optical paths differ bya multiple of , the reflectedwaves add. If the paths cause a phase difference , reflected waves cancel out.

Thin Film Interference

1

t

2oil on wateroptical film on glasssoap bubble

n = 1

n > 1

Ray 1 has a phase change of phase of upon reflectionRay 2 travels an extra distance 2t (normal incidence approximation)

Constructive interference: rays 1 and 2 are in phase

2 t = m n + ½ n 2 n t = (m + ½) [n = /n]

Destructive interference: rays 1 and 2 are out of phase

2 t = m n 2 n t = m

Thin Film Interference

Thin films work with even lowcoherence light, as paths are short

Different wavelengths will tend to add constructively at different angles, and we see bands of different colors.

When ray 2 is in phase with ray 1, they add up constructively and we see a bright region.

When ray 2 is out of phase, the rays interfere destructively.This is how anti-reflection coatings work.

1

t

2oil on wateroptical film on glasssoap bubble

n = 1

n > 1

Diffraction

What happens when a planar wavefront

of light interacts with an aperture?

If the aperture is large compared to the wavelengththis would be expected....

…Light propagating in a straight path.

If the aperture is small compared to the wavelength,would this be expected?

Diffraction

Not really…

In fact, what happens is that:

a spherical wave propagates out from the aperture.

All waves behave this way.

Diffraction

This phenomenon of light spreading in a broad pattern, instead of following a straight path, is called: DIFFRACTION

If the aperture is small compared to the wavelength,would the same straight propagation be expected? … Not really

Angular Spread: ~ / a

Slit width = a, wavelength =

I

Diffraction

Huygen’s Principle

Huygen first explained this in 1678 by proposing that all planar wavefronts are made up of lots of spherical wavefronts..

Huygen’s Principle

Huygen first explained this in 1678 by proposing that all planar wavefronts are made up of lots of spherical wavefronts..

That is, you see how light propagatesby breaking a wavefront into littlebits

Huygen’s Principle

Huygen first explained this in 1678 by proposing that all planar wavefronts are made up of lots of spherical wavefronts..

That is, you see how light propagatesby breaking a wavefront into littlebits, and then draw a spherical waveemanating outward from each littlebit.

Huygen’s Principle

Huygen first explained this in 1678 by proposing that all planar wavefronts are made up of lots of spherical wavefronts..

That is, you see how light propagatesby breaking a wavefront into littlebits, and then draw a spherical waveemanating outward from each littlebit. You then can find the leading edge a little later simply by summing allthese little “wavelets”

Huygen’s Principle

Huygen first explained this in 1678 by proposing

that all planar wavefronts are made up of lots of

spherical wavefronts..

That is, you see how light propagatesby breaking a wavefront into littlebits, and then draw a spherical waveemanating outward from each littlebit. You then can find the leading edge a little later simply by summing allthese little “wavelets”

It is possible to explain reflection and refraction this way too.

Diffraction at Edges

what happens to the shape of the field at thispoint?

Diffraction at Edges

Light gets diffracted at the edge of an opaque barrier there is light in the region obstructed by the barrier

I

Double-Slit Interference

Because theyspread, these waves willeventuallyinterfere withone another andproduceinterferencefringes

Double-Slit Interference

Double-Slit Interference

Double-Slit Interference

Double-Slit Interference

Brightfringes

Double-Slit Interference

Brightfringes

screen

Double-Slit Interference

Brightfringes

Thomas Young (1802) used double-slit interference to prove the wave nature of light.

screen

L

d

y r1

r2

P

Double-Slit Interference

Light from the two slits travels different distances to the screen.The difference r1 - r2 is very nearly d sin. When the path difference is a multiple of the wavelength these add constructively,and when it’s a half-multiple they cancel.

y

d sin = m bright fringes

d sin = ( m+1/2) dark fringes

Now use y = L tan ; and for small y sin tan = y / L

y bright = mL/d

y dark = (m+ 1/2)L/d L

d

r1

r2

P

Double-Slit Interference

Light from the two slits travels different distances to the screen.The difference r1 - r2 is very nearly d sin . When the path difference is a multiple of the wavelength these add constructively,and when it’s a half-multiple they cancel.

With more than two slits, things get a little more complicated

L

d

y

P

Multiple Slit Interference

Now to get a brightfringe, many pathsmust all be in phase.

The brightest fringesbecome narrower butbrighter;

and extra lines show upbetween them.

Multiple Slit Interference

With more than two slits, things get a little more complicated

L

d

y

P

Such an array of slits is called a “Diffraction Grating”

The most intense diffraction lines appear when:

d sin() = m

Note that each wavelength is diffractedat a different angle

S

Multiple Slit Interference

All of the lines (more intense and less intense) show up at the set of angles given by: d sin = (n/N) (N = number of slits).

Single Slit Diffraction

Angular Spread: ~ / a

IIntensity

distribution

Each point in the slit acts as a source of spherical wavelets

Slit width a

For a particular direction , wavelets will interfere, either constructively or destructively,resulting in the intensity distribution shown.

The Diffraction Limit

Diffraction imposes a fundamental limit on the resolution of optical systems:

Suppose we want to image 2 distant points, S1 and S2, through an aperture of width a:

Two points are resolved whenthe maximum of one is at

the minimum of the second

The minima occurs forsin = / a

a = slit width

Using sin min = / a

S1

S2

L

D

Dmin / L / a

Example: Double Slit Interference

50 cm

d

Light of wavelength = 500 nm is incident on a doubleslit spaced by d = 50 m. What is the fringe spacing onthe screen, 50 cm away?

Example: Double Slit Interference

50 cm

d

Light of wavelength = 500 nm is incident on a doubleslit spaced by d = 50 m. What is the fringe spacing onthe screen, 50 cm away?

yL / d

50 10 2m 500 10 9m 50 10 6m 5mm

Example: Single Slit Diffraction

a

50 cm

Light of wavelength = 500 nm is incident on a slit a=50 m wide. How wide is the intensity distribution on the screen, 50 cm away?

Example: Single Slit Diffraction

a

50 cm

Light of wavelength = 500 nm is incident on a slit a=50 m wide. How wide is the intensity distribution on the screen, 50 cm away?

/a

yL 50 10 2m 500 10 9m 50 10 6m 5mm

What happens if the slit width is doubled?The spread gets cut in half.

Beams of green ( = 520 nm) and red ( = 650 nm) lightimpinge normally on a grating with 10000 lines per cm.

What is the separation y of the first order diffracted beamson a screen, parallel to the grating, and located at a distanceL = 50 cm away from it?

Grating equation d sin = m

sin y/L for small angles, and m = 1 for first order10000 lines/cm d = 0.0001

d y / L = y = L /d

Y green = 520x10-7x50 / 0.0001 = 26 cmY red = 650x10-7x50 / 0.0001 = 32.5 cm

y = 6.5 cm