Diffraction

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Chapter 25: Interference and Diffraction

•Constructive and Destructive Interference

•The Michelson Interferometer

•Thin Films

•Young’s Double Slit Experiment

•Gratings

•Diffraction

•Resolution of Optical Instruments

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Interference and DiffractionCertain physical phenomena are most easily explained by invoking the wave nature of light, rather than the particle nature.

These include:interference, diffraction, polarization

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CoherenceIf the phase of a light wave is well defined at all times (oscillates in a simple pattern with time and varies in a smooth way in space at any instant), then the light is said to be coherent.If, on the other hand, the phase of a light wave varies randomly from point to point, or from moment to moment (on scales coarser than the wavelength or period of the light) then the light is said to be incoherent.For example, a laser produces highly coherent light. In a laser, all of the atoms radiate in phase.An incandescent or fluorescent light bulb produces incoherent light. All of the atoms in the phosphor of the bulb radiate with random phase. Each atom oscillates for about 1ns, and produces a wave about 1 million wavelengths long.

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Interference

Just like sound waves, light waves also display constructive and destructive interference.

For incoherent light, the interference is hard to observe because it is “washed out” by the very rapid phase jumps of the light.Soap films are one example where we can see interference effects even with incoherent light.

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Soap Film Interference (White light)

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Conditions for Interference

The following four conditions must be true in order for an interference pattern to be observed.

The source must be coherent – has a constant phase relationship

Wavelengths must be the same –monochromatic

The Principle of Superposition must apply.The waves have the same polarization state.

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Constructive and Destructive Interference(the same for all waves!)

Two waves (top and middle) arrive at the same point in space.

The total wave amplitude is the sum of the two waves.

The waves can add constructively or destructively

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Constructive interference occurs when two waves are in phase. To be in phase, the points on the wave must have Δφ=(2π)m, where m is an integer.

When coherent waves are in phase, the resulting amplitude is just the sum of the individual amplitudes. The energy content of a wave depends on A2. Thus, I∝A2.

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The resulting amplitude and intensity are:

2121

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2 IIIII

AAA

++=

+=

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Destructive interference occurs when two waves are a half cycle out of phase. To be out of phase the points on the wave must have Δφ=(2π)(m+½), where m is an integer.

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The resulting amplitude and intensity are:

2121

21

2 IIIII

AAA

−+=

−=

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Coherent waves can become out of phase if they travel different distances to the point of observation.

This represents the extra path length (Δl) that the wave from S2 must travel to reach point P.

θsindl =Δ

P

d

S2

S1

θ

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For constructive interference λml =Δ

where m = an integer.

For destructive interference λ⎟⎠⎞

⎜⎝⎛ +=Δ

21ml

where m = an integer.

When both waves travel in the same medium the interference conditions are:

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FringesIf at point P the path difference yields a phase difference of 180 degrees between the two beams a “dark fringe” will appear. If the two waves are in phase, a “bright fringe” will appear.

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Example A 60.0 kHz transmitter sends an EM wave to a receiver 21 km away. The signal also travels to the receiver by another path where it reflects from a helicopter. Assume that there is a 180° phase shift when the wave is reflected.

(a) What is the wavelength of this EM wave?

km 0.5Hz 1060

km/sec 100.33

5

=×

×==

fcλ

(b) Will this situation give constructive interference, destructive inference, or something in between?

The path length difference is Δl = 10 km = 2λ, a whole number of wavelengths. Since there is also a 180° phase shift there will be destructive interference.

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Michelson Interferometer

In the Michelson interferometer, a beam of coherent light is incident on a beam splitter. Half of the light is transmitted to mirror M1 and half is reflected to mirror M2.

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The beams of light are reflected by the mirrors, combined together, and observed on the screen.

If the arms are of different lengths, a phase difference between the beams can be introduced.

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Example A Michelson interferometer is adjusted so that a bright fringe appears on the screen. As one of the mirrors is moved 25.8 μm, 92 bright fringes are counted on the screen. What is the wavelength of the light used in the interferometer?

Moving the mirror a distance d introduces a path length difference of 2d. The number of bright fringes (N) corresponds to the number of wavelengths in the extra path length.

m 561.022

μλ

λ

==

=

Nd

dN

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Thin FilmsWhen an incident light ray reflects from a boundary with a higher index of refraction, the reflected wave is inverted (a 180° phase shift is introduced).

A light ray can be reflected many times within a medium.

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Phase of wave reflected by interface between two media

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Interference in Thin FilmsWe have all seen the colorful patterns which appear in soap bubbles. The patterns result from an interference of light reflected from both surfaces of the film. Some colors undergo constructive some destructive interference:

0o phase change

tn>1

180o phase change

2t = mλn = mλ / n (m = 0, 1, 2…) Destructive (invisible)

2t = (m+1/2)λn = (m + ½)λ / n (m = 0, 1, 2…) Constructive (bright color)

λ / n= λn

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Air WedgeMonochromated Light

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Newton’s Rings

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Example A thin film of oil (n=1.50) of thickness 0.40 μm is spread over a puddle of water (n=1.33). For which wavelength in the visible spectrum do you expect constructive interference for reflection at normal incidence?

AirWater Oil

Incident wave

Consider the first two reflected rays. r1 is from the air-oil boundary and r2 is from the oil-water boundary.

r1 has a 180° phase shift (noil >nair), but r2 does not (noil<nwater).

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To get constructive interference, the reflected waves must be in phase. For this situation, this means that the wave that travels in oil must travel an extra path equal to multiplesof half the wavelength of light in oil.

The extra path distance traveled is 2d, where d is the thickness of the film. The condition for constructive interference here is:

⎟⎠⎞

⎜⎝⎛ +

=∴

⎟⎟⎠

⎞⎜⎜⎝

⎛⎟⎠⎞

⎜⎝⎛ +=⎟

⎠⎞

⎜⎝⎛ +=

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221

212

oilair

oil

airoil

m

dnn

mmd

λ

λλ

Only the wavelengths that satisfy this condition will have constructive interference.

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Make a table:

0.2740.3430.4820.8012.400

λair(μm)m

All of these wavelengths will show constructive interference, but it is only this one that is in the visible portion of the spectrum.

Example continued:

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White light is incident on a soap film (n = 1.30) in air. The reflected light looks bluish because the red light (λ = 670 nm) is absent in the reflection. What is the minimum thickness of the soap film?

2t = mλ / n (m = 1, 2…) Destructive interference

t = 1λ / 2n = 258nm

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Young’s Double-Slit Experiment

Place a source of coherent light behind a mask that has two vertical slits cut into it. The slits are L tall, their centers are separated by d, and their widths are a.

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The slits become sources of waves that, as they travel outward, can interfere with each other.

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The pattern seen on the screen

There are alternating bright/dark spots.

An intensity trace

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If the two slits are separated by a distance d and the screen is far away then the path difference at point P is Δl = dsinθIf Δl = λ, 2λ, 3λ, etc, then the waves will arrive in phase and there will be a bright spot on the screen.

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Interference ConditionsFor constructive interference, the path difference must be zero or an integral multiple of the wavelength:

For destructive interference, the path difference must be an odd multiple of half wavelengths:

n is called the order number

2...) 1, 0,( , = sin ±±=nnd λθ

2...) 1, 0,( 1/2( = sin ±±=+ nnd λθ )

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Example: Show that the interference fringes in a double-slit experiment are equally spaced on a distant screen near the center of the interference pattern.

The condition for constructive interference is

dm

mdlλθ

λθ

=∴

==Δ

sin

sin

From the geometry of the problem,

Dh

=θtan

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The screen is far away compared to the distance between the slits (D>>d) so tanθ ≈ sinθ ≈ θ. Here,

dDmh

Dh

dm

Dh

dm

λ

λ

θθλθθ

=

=∴

=≈=≈ tan and sin

The distance between two adjacent minima is:

( )dDmm

dDhh λλ

=−=− 1212

Example continued:

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ExampleIf the distance between two slits is 0.050 mm and the distance to a screen is 2.50 m, find the spacing between the first- and second-order bright fringes for yellow light of 600 nm wavelength.

cmmm

mmdLdLxx

606.01005.0

5.2106002//2

3

9

12

==⋅

⋅⋅⋅=

−≈−

−

−

λλ

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When green light (λ = 505 nm) passes through a pair of double slits, the interference pattern shown in (a) is observed. When light of a different color passes through the same pair of slits, the pattern shown in (b) is observed. (a) Is the wavelength of the second color greater than or less than 505 nm? Explain. (b) Find the wavelength of the second color. (Assume that the angles involved are small enough to set sinθ = tanθ.)

nm

dLdLxx

green

greengreen

45555.4

/5/5.4

?

?5?,5.4,

==⇒

=⇒=

λλ

λλ

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Gratings

A grating has a large number of evenly spaced, parallel slits cut into it.

λθ md =sin

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Example Red light with λ=650 nm can be seen in three orders in a particular grating. About how many rulings per cm does this grating have?

For each of the maxima

λθλθλθ

λθθ

4sin3sin2sin1sin0sin

4

3

2

1

0

=====

ddddd

This order is not observed.

Third order is observed.

λθ md =sin

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Since the m = 4 case is not observed, it must be that sinθ4>1. We can then assume that θ3≈90°. This gives

m 1095.13 6−×== λd

and lines/cm. 5100lines/m 000,5101===

dN

Example continued:

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Diffraction

Using Huygens’s principle: every point on a wave front is a source of wavelets; light will spread out when it passes through a narrow slit.

Diffraction is appreciable only when the slit width is nearly the same size or smaller than the wavelength.

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Definition and Types of DiffractionDefinition and Types of Diffraction•• Diffraction is the bending of a wave around an Diffraction is the bending of a wave around an

object accompanied by an interference patternobject accompanied by an interference pattern•• Fresnel DiffractionFresnel Diffraction -- curved (spherical) wave curved (spherical) wave

front is diffractedfront is diffracted•• Fraunhofer DiffractionFraunhofer Diffraction -- plane wave is diffractedplane wave is diffracted

Fresnel Diffraction fromFresnel Diffraction froma circular obstructiona circular obstructionFresnel bright spotFresnel bright spot

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The intensity pattern on the screen.

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λθ ma =sinThe minima occur when:

where m = ±1, ±2,…

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Example: Light from a red laser passes through a single slit to form a diffraction pattern on a distant screen. If the widthof the slit is increased by a factor of two, what happens to thewidth of the central maximum on the screen?

a

maλθ

λθ

≈

=sin

The central maximum occurs between θ=0 and θ as determined by the location of the 1st minimum in the diffraction pattern:

Let m =+1 and assume that θ is small.

From the previous picture, θ only determines the half-width of the maximum. If a is doubled, the width of the maximum is halved.

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Resolution of Optical Instruments

The effect of diffraction is to spread light out. When viewing two distant objects, it is possible that their light is spread out to where the images of each object overlap. The objects become indistinguishable.

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Rayleigh’s Criterion: two point sources are barely resolvable if their angular separation θR results in the central maximum of the diffraction pattern of one source’s image is centered on the first minimum of the diffraction pattern of the other source’s image.

Resolvability

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Fig. 36-10

small1sin 1.22 1.22 (Rayleigh's criterion)

R

R d d

θλ λθ − ⎛ ⎞= ⎜ ⎟⎝ ⎠ ≈

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For a circular aperture, the Rayleigh criterion is:

λθ 22.1sin ≥Δa

where a is the aperture size of your instrument, λ is the wavelength of light used to make the observation, and Δθ is the angular separation between the two observed bodies.

Δθ

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λθ 22.1sin ≥ΔaTo resolve a pair of objects, the angular separation between them must be greater than the value of Δθ.

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Example: The radio telescope at Arecibo, Puerto Rico, has a reflecting spherical bowl of 305 m diameter. Radio signals can be received and emitted at various frequencies at the focal point of the reflecting bowl. At a frequency of 300 MHz, what is the angle between two stars that can barely be resolved?

( )( )

degrees

m Hz 10300

m/s 6

23.0101.4sin

305

100.322.122.1sin

22.1sin

3

8

≥Δ×≥Δ

××

=≥Δ

≥Δ

−

θθ

λθ

λθ

a

a

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X-rays are electromagnetic radiation with wavelength ~1 Å = 10-10 m (visible light ~5.5x10-7 m)

X-Ray Diffraction

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Fig. 36-27

X-ray generation

X-ray wavelengths to short to be resolved by a standard optical grating

( )( )1 1 1 0.1 nmsin sin 0.0019

3000 nmmdλθ − −= = = °

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Diffraction of x-rays by crystal: spacing d of adjacent crystal planes on the order of 0.1 nm

→ three-dimensional diffraction grating with diffraction maxima along angles where reflections from different planes interfere constructively

X-Ray Diffraction, cont’d

36-Fig. 36-28

2 sin for 0,1, 2 (Bragg's law)d m mθ λ= = …