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ANNOUNCEMENT •  Final: Wednesday, December 14, 2016, 7 PM - 9 PM •  Location: Elliot Hall of Music •  Covers all readings, lectures, homework from

Chapters 28.4 through 33 –  Multiple choice

•  Practice exams on the course website and on CHIP •  Bring your student ID card and your own 1-page

(8-1/2” x 11”, both sides) crib sheet plus the original (or revised) crib sheets that you prepared for Exams 1 and 2 –  Only a few equations will be given

•  The equation sheet that will be given with the exam is posted on the course homepage (“Equationsheet”)

–  It is your responsibility to create your own crib sheets

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LECTURE 27: Interference, Diffraction, Resolution

Interference Different Path Lengths

•  The phase difference between two waves can change if the waves travel paths of different lengths

•  Interference maxima

•  Interference minima

d sinθm = mλm = order number = 0,1,2,...

d sinθm = m− 12

⎛

⎝⎜⎜⎜

⎞

⎠⎟⎟⎟λ

m = 1,2,3,...

3

Interference Different Path Lengths

I=4I0 cos2 δ

2δ= 2πd

λsinθ

Diffraction •  Geometrical optics

–  When length scales of an edge, obstacle, or aperture are >> than the light’s wavelength

•  Light propagates as rays •  Forms bright, sharp image and dark, sharp shadows

•  Wave optics –  When length scales of an edge, obstacle, or aperture are

comparable to the light’s wavelength •  Waves spread out

–  Can interfere constructively or destructively

= Diffraction •  Interference patterns of light and dark fringes

–  Geometric optics not a good approximation

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Diffraction •  Consider the case of light impinging on a small disk

We observe: –  Bright spot in the center –  Diffraction rings outside and inside the geometrical shadow area

•  The bright spot at the center was predicted by Fresnel in 1818 and observed by Arago

Diffraction from a Disk

•  Newton was a proponent of geometric optics (black lines)

•  Fresnel believed in wave optics (red lines) –  Waves that touch the top and bottom of the disk

travel the same distance to the center •  Interfere constructively

Intensity

Cyl

indr

ical

ob

stac

le

Plane wave sc

reen

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Diffraction from an Edge

Diffraction pattern

Plane wave

Obstacle

Diffraction from a Single Slit (screen far away)

•  Consider a monochromatic wave incident on a narrow slit

•  Geometrical optics predicts that the transmitted beam has the same cross section as the slit

•  Experiments show that wave optics is correct –  Central bright band that is wider than the width of the slit

–  Alternating dark and bright fringes border the central bright band

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Diffraction from a Single Slit

•  Central bright fringe –  Waves from all

points in the slit travel the same distance to reach the center

–  => are in phase

When 12

asinθ= 12λ,

destructive interference betweenthe paired rays occurs

Diffraction from a Single Slit

•  Represent the slit as a number of point sources of equal amplitude

•  Divide the slit into two •  Pair a point from the upper half

with its partner in the lower half

a2

sinθ

(first minimum)

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Diffraction from a Single Slit

asinθ1 =λ (location 1st min)

asinθ2 = 2λ (location 2nd min)

Locations of minima for single-slit diffraction:asinθm = mλ, m = 1,2,3,...

tanθm =

ym

L

Diffraction from a Single Slit

Incr

easi

ng a

pert

ure

Dependence of the central maximum on a:

a sinθ= mλ→ asinθ=λ

sinθ≈θ= λa

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Question Two wavelengths, 650 and 430 nm, are used separately in a single-slit diffraction experiment. The figure shows the results as graphs of intensity, I, versus angle q for two diffraction patterns. If both wavelengths are then used simultaneously, what color will be seen in the combined diffraction pattern at angle D?

I

q D E

(A)  Violet (B)  Red

DEMO Diffraction from a Circular Aperture

•  The diffraction pattern of a circular aperture of diameter d is similar to a single slit of width a

•  Airy disk •  Central bright spot

•  About 85% of the power is in this area

•  The dark fringes are found at:

sinθ1 = 1.22λd

sinθ2 = 2.23λd

sinθ3 = 3.24λd

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•  The bright fringes are at:

•  The Airy disk limits the resolvability of nearby objects

Diffraction from a Circular Aperture

Image of two nearby binary stars but diffraction patterns overlap

sinθ1 = 1.63λd

sinθ2 = 2.68λd

sinθ3 = 3.70λd

Rayleigh Criterion When 1.22C d

≈

1

2

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Maximum Falls on Minimum •  The minimum angular separation αc of two marginally

resolvable points –  Maximum of the diffraction pattern from one falls on the first

minimum of the diffraction pattern of the other

•  The first minimum is at αC ≈1.22λ

d

Cen

tral A

xis

Rayleigh Criterion •  The minimum angular separation αc of two marginally

resolvable points –  Maximum of the diffraction pattern from one falls on the first

minimum of the diffraction pattern of the other

•  The first minimum is at

•  Therefore

αC ≈1.22λ

d

αC = θ=

sin−1 1.22λd

⎛

⎝⎜⎜⎜

⎞

⎠⎟⎟⎟≈1.22λ

d

Not resolved Resolved Barely resolved

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Rayleigh Criterion •  If α> αC, objects can be resolved •  If α< αC, objects cannot be resolved •  To increase our ability to distinguish objects we

must minimize the diffraction pattern

–  αC also depends on relative brightness and atmospheric turbulence

–  Can increase d or decrease λ

•  Use ultraviolet light

•  e- beam used in Scanning Electron Microscopes (SEM) have λ≈λ(light)/105

•  Place object under a microscope in a drop of oil λo=λ/n

αC ≈1.22λ

d

Diffraction Gratings •  What happens if we go from 2 slits to N slits? •  The fringes become narrower and faint secondary

maxima appear between the fringes –  Half-width of central line

•  If N is large (N/ℓ≈104/cm) the fringes are very sharp –  Secondary maxima can be neglected and you have a grating

d δ

θ θ

d = gratingspacing

rulings⇒ N(lines)cm

= 5000 linescm

d = cmN

d = 15000⎛

⎝⎜⎜⎜

⎞

⎠⎟⎟⎟cm = 2×10−4cm

δ= path length difference = d sinθ

Δθhw = λ

Nd

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DEMO Diffraction Gratings

•  Sharp bright fringes occur if

–  where m is the order of the maxima •  Gratings are used to measure λ

–  by detecting maxima of the diffraction pattern with m=1,2,…

•  Resolving power

δ= d sinθ= mλ m = 0,1,2

sinθ= mλ

d

R = λ

Δλ= mN

Chose theta (hence m) and d, solve for lambda

In demo: wavelength is fixed, d decreases as N increases…so spacing between fringes increases…illumination region from laser is constant

Diffraction Gratings •  For a given

wavelength and d, if N increases the half-width decreases

•  Fixed d &

wavelength –  Changing the

aperture to illuminate more lines

–  Different than the demo

Δθhw = λ

Nd