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PHYS241 - Physics - Purdue University

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1 Please fill out online course evaluation http://www.purdue.edu/eval Dr. Pyrak-Nolte and I would really appreciate your feedback THANKS! 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
Transcript

1

Please fill out online course evaluation

http://www.purdue.edu/eval

Dr. Pyrak-Nolte and I would really appreciate your feedback

THANKS!

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

2

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

4

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

5

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

6

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)

7

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

8

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

9

•  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

10

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

11

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

12

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


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