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Chapter 4 Diffraction of Light In Chapter 3, we saw how light beams passing through different slits can interfere with each other and how a beam after passing through a single slit flares-diffracts- in Young's experiment. Diffraction through a single slit or past either a narrow obstacle or an edge produces rich interference patterns. The physics of diffraction plays an important role in many scientific and engineering fields. In this chapter we explain diffraction using the wave nature of light and discuss several applications of diffraction in science and technology.

•  Single Slit Diffraction •  Rayleigh’s Criterion •  Diffraction Grating •  X-ray Diffraction

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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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Diffraction

Diffraction is the scattering of light from objects or orifices. For plane waves entering the single slit, the waves emerging from the slit start spreading out, or diffracting.

Diffractive Scattering ~ λ/a

λ < a λ = a λ > a

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Hold your thumb and index finger very close together, but without quite closing the gap. With one eye, look through the gap at a bright background or a light source - your monitor will do nicely. With the right separation, you’ll see one or more dark bands appear in the narrow gap. Move your fingers up close to the eye - the effect you’re looking for will be much more pronounced. You’re witnessing Fresnel diffraction from a single slit.

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Single Slit Diffraction

IP(θ ) = 1µ0c

|!EP |2= 1

µ0c|

n∑!En |2 →

!EP ~ E0 s in(

!k ⋅ !r −ωt +φ(!r )]dy

−a/2

+a/2

∫!EP = E0

sinαα

where α = π asinθλ

P

!r

θ a2

a2

IP θ( ) = I0

sinαα

⎛⎝⎜

⎞⎠⎟

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α = mπ m = 1,2,3.... min ima

α = π asinθλ

→ asinθ = mλ m = 1,2,3... min ima

L a

sinθθ!θ − θ

3

3!+ θ

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5!+ ...

θ= 1− θ

2

6+ ...

I m = 0( ) = I0

(max) θ = 0 I m = 1( ) = 0 (min) θ = + λ

a

I m = 1( ) = 0 (min) θ = − λa

I m = 2( ) = 0 (min) θ = − 2λa

I m = 2( ) = 0 (min) θ = + 2λa

λ

I0

I m = 3( ) = 0 (min) θ = + 3λa

I m = 3( ) = 0 (min) θ = − 3λa

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Measuring the size of small objects with diffraction

Diffractive scattering from a slit and target object are nearly identical. A bright diffractive maxima appears behind the object. We can use the dark fringe pattern to estimate the size of the object

Q1: Laser light of λ=650nm is incident on a small object. The width of the central maxima is measured to be 2.6 cm. The screen is 2 m from the cylinder.

λ=650nm

sinθ = mλa

position of min ima θ = mλa

, θ <<1

Δθ = 2 λa

acrosscentral min ima

Δy = LΔθ = 2Lλa

a = 2LλΔy

a = object size

a = 2(2m) (650nm)2.6cm

= 0.0001m = 100µm

https://www.youtube.com/watch?v=kpsN78mQ6YY

L

Δθ Δy a

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Diffraction by a Circular Aperture a

sinθm=1 = 1.22 λa

(1st min.- circ. aperture)

sinθm=1 =λa

(1st min.- single slit)

Light Light a

θ θ

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98%

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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 of Light Sources

θR = sin−1 1.22

λd

⎛⎝⎜

⎞⎠⎟≅ 1.22

λn

d (Rayleigh's criterion)

θ = θR θ >θR θ <θR

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Calculate angle θ = tan−1(ΔxL

) ! ΔxL

L !Δxθ

where θ = θR at the lim it

L

Δx

θR

θ

Resolving distant light sources or structures

Q2: A car’s headlights are separated by about 1.3m. At what distance L, based on the Rayleigh criteria, will a person be able to resolve the car’s headlights as being separated and not from a single light source, e.g. coming from a motorcycle? Assume yellow light has wavelength λ=560nm. Pupil dia~4mm.

D ~ 4mm pupil diameter

θR = 1.22 λD

RayleighCondition

= 1.22 (560nm)4mm

= 1.7 ×10−4

θ = θR

L = ΔxθR

= 1.3m1.7 ×10−4 = 7650m

Diffraction by a Double Slit

•  Diffraction from a double slit. The purple line with peaks of the same height are from the interference of the waves from two slits; the blue line with one big hump in the middle is the diffraction of waves from within one slit; and the thick red line is the product of the two, which is the pattern observed on the screen. The plot shows the expected result for a slit width D = 2𝜆 and slit separation d = 6𝜆. The maximum of m = ±3 order for the interference is missing because the minimum of the diffraction occurs in the same direction.

The intensity pattern from a Double slit Experment will be a Superposition of Interference And Diffraction patterns. Since θd~λ/a for diffraction, for a~λ the effect is most pronounced.

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(a) Light passing through a diffraction grating is diffracted in a pattern similar to a double slit, with bright regions at various angles.

(b) The pattern obtained for white light incident on a grating. The central maximum is white, and the higher-order maxima disperse white light into a rainbow of colors.

Diffraction Gratings

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Light of wavelength λ projected on to an N slit grating constructively interfere and form a diffraction maxima at multiples of mλ. Light of many wavelengths will separate in to spectra at the m=±1, ±2,… orders . At the central m=0 maxima all optical paths coincide producing no spectral separation.

Diffraction Gratings

d sinθ = mλ for m = 0,1,2… (maxima-lines)

d sinθ

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Separates different wavelengths (colors) of light into distinct diffraction lines. Grating Spectroscope

Spectrometer Dispersion DiDispersion describes the spreading of the lines

in angle per unit wavlength D = dθdλ

≅ md

d sinθ = mλ → d cosθ Δθ = mΔλ

D = ΔθΔλ

= md cosθ

~ md

θ <<1

LightSource

Wavelength (nm)

Hg Lamp Spectrum

iDispersion increases with order m and decreases with spacing d

ResolvingPower RiResolving power desribes how well two lines in a spectrometer can be resolved. iAs we increase the number of lines of the grating N the resolution will improve,

with Δλ ~ 1/ N . The resolving power is then proportional to N: R = λΔλ

~ N

iThe resolving power also increases with order mof the grating. We then conclude that

the resolving power of a grating is: R = λΔλ

= mN e.g. R = 1000 = 1×1000, 2×500,....

i Note that R is independent of d andλ.

R=200 R=500 R=1000

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Q3: Consider a diffraction grating with N = 500 and d = 4000 nm, used to view the spectral lines of the sodium doublet, two closely spaces lines in the sodium spectrum. The sodium doublet has λ = 589.00 nm, and Δλ = 0.59 nm. (a) Find the resolving power needed to resolve the sodium doublet. (b) Find the resolving power R of the grating, for m = 1. (a) R = λ/Δλ = 589/0.59 = 1000 resolving power needed. (b) R = Nm = 500 is the resolving power of the grating with N=500, m=1.

•  The two lines will not be able to be resolved with an N=500 grating in first order!

•  If m=2 selected, R can be increased to 1000, making the separation just resolvable.   

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•  X-rays are electromagnetic radiation with wavelengths 10-11 - 10-7 m.

•  These X-rays typically have energies in the 1-100 KeV range.

X-Ray Diffraction for resolving atomic/molecular distances

X-ray generation

•  X-ray are generated by accelerating electrons in to a heavy metal target, like tungsten, W.

•  E(eV)=1240/λ(nm) e.g. 60KeV->λ=2e-11m

100kV

K

L

M

lines

Continuous xrays from electron deceleration in W

λ(nm) = 1240

E(eV )

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

•  Diffraction maxima occur along angles where reflections from different planes interfere constructively

•  Braggs Law can be used to analyse X-ray diffraction patterns revealing the structure of crystals and molecule structure.

•  The structure of DNA was discovered largely in part by X-ray diffraction experiments conducted by British chemist and crystallographer Rosalind Franklin.

X-Ray Diffraction, cont’d

Bragg's law2d sinθ = mλ for m = 1,2,3… (max ima)

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Ray-2 will travel an extra distance δ=2dsinθ w.r.t. ray-1 creating a phase shift Δφ. If the phase shift Δφ = mλ, m = 1,2,3.., there will be constuctive interference.

1

2

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Q4: In a SiC crystal a first-order maximum is observed at an incidence angle of 27.1°. The Xray energy used in the scan 5keV. What is the crystal lattice spacing d being observed?

Bragg's law2d sinθ = mλ for m = 1,2… (max ima)

d = mλ2sinθ

= (1)(0.25nm)2sin(27.10 )

= 0.27nm

https://serc.carleton.edu/NAGTWorkshops/deepearth/activities/40414.ht

Wavelength

λ(nm) = 1240eV − nmE(eV )

λ = 12405,000eV

nm = 0.25nm

m=1

m=1

m=1

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STOP HERE