Chapters 3638 Review

1
Chapters 36-38 Review
PHYS 2326-36
2
Suggestions

• Take time to create your note card
• It can be a full sized sheet containing formulas
  and constants but no worked problems or problem
  solutions
• Know what is on the sheet and where
• Do not put things you dont know how to use and
  dont put too much on there
• Organize it and label it
• Include your name and turn it in with the test
  you will probably want it back for the final

3
Test 5

• Test 5 covers Light, lecture notes 30-35 and
  chapters 35-38
• You are responsible for knowing the contents in
  those chapters
• If it wasnt discussed in class, it might not be
  on the test
• If it was discussed in class, it may well be on
  the test
• The lecture notes include most of what was
  discussed in class

4
Chapter 35 Light
PHYS 2326-30
5
Concepts to Know

• Speed of Light
• Wave Fronts Rays
• Reflection
• Law of Reflection
• Refraction
• Index of Refraction
• Law of Refraction (Snells Law)
• Total Internal Reflection

6
Concepts to Know

• Critical Angle
• Huygens Principle
• Dispersion

7
Speed of Light

• Light is very fast
• 2.998 x 10 8 m/s in a vacuum (3.0E8 )
• slower in materials such as air or glass

8
Rays

• Can assume light travels in fixed directions in a
  straight line when passing through a uniform
  medium
• This is the ray approximation
• Rays are perpendicular to wavefronts
• Works well for studying mirrors, lenses, prisms
  and optical instruments like telescopes
• Light ray paths are reversible

9
Reflection

• Like waves traveling along a string hitting a
  boundary, light can reflect when hitting a
  boundary
• Unlike a string, light doesnt have to reflect
  back along the same direction
• Specular reflection is from a smooth surface like
  a mirror or a calm pond
• Diffuse reflection occurs from rough surfaces

10
Law of Reflection

• Given a smooth surface at a point, there is a
  normal to that surface
• Reflection occurs so that the angle of the
  reflected ray is the same as the angle of the
  incoming or incident ray both measured from the
  normal

?
?
11
Refraction

• When a boundary exists between two different
  transparent media, the light ray passing through
  the boundary is bent and is said to be refracted,
  depending upon the properties of the two media
• The relationship for this is eqn 35.3
• where ?1 and v 1 are the angle of incidence and
  speed of light in the first medium and ?2 and v2
  are for the second medium

12
Refraction Reflection

• Usually, both occur at a boundary
• When going from air to a denser material with a
  lower speed of light water, glass refraction
  bends the ray towards the normal
• When going from denser material to air, the ray
  bends away from the normal

N
?1
?1
?2
13

• Example of light going through a glass to air
• As light travels from one medium to another, the
  frequency doesnt change but the wavelength and
  velocity change

?2
N
?1
?1
?1
?2
14
Index of Refraction

• n defined as c/v the ratio of the speed of
  light in a vacuum divided by the speed of light
  in the medium

15
Refractive Index

• Table 35.1 shows a refractive index for various
  materials
• A vacuum is 1.00000000
• Air is close at 1.000293
• Water is just over 1.3
• Glass varies around 1.4 to 1.7

16
Huygenss Principle

• All points on a given wave are taken as point
  sources for the production of spherical secondary
  waves called wavelets

17
Dispersion

• The index of refraction is not constant with
  wavelength
• Dispersion is the behavoir where the angle of
  refraction changes with wavelength
• This creates the rainbow effect of rain droplets
  and prisms

18
Total Internal Reflection

• Note that a sin function varies from -1 to 1 and
  that the typical index of refraction n can vary
  from 1.0 to over 2.0.
• For a surface boundary between materials 1 and 2,
  where material 1 has a greater index of
  refraction than material 2, what happens when the
  angle ?2 reaches 90 degrees?

19

• n1 sin ?1 n2 sin 90 n2
• ?1 becomes the critical angle ?c
• sin ?1 n2 / n1 (for n1 gt n2) eqn 35.10
• As the incident angle ?1 reaches the critical
  angle ?c there is total internal reflection going
  on as all light is reflected since the refraction
  angle cannot exceed 90.

gt?c
20
Chapter 36 Mirrors and Image Formation
PHYS 2326-31
21
Concepts to Know

• Mirrors
• Plane, Concave, Convex
• Sign Conventions
• Object Distance
• Image Distance
• Focal Length
• Magnification

22
Concepts to Know

• Enlarged, Reduced
• Erect, Inverted
• Real, Virtual
• Reversed (Perverted)
• Converging, Diverging
• Aberration
• Principle Rays

23
Definitions

• Plane surface one that is flat
• Concave surface, one that is curved inward at the
  center
• Convex surface, one that is curved outward

Eye
Plane
Concave
Convex
24
Objects Images

• Image from a Plane Mirror
• p object to surface, q surface to image
• h object height, h image height
• Magnification M 1 for flat mirror

p
q
h
h
?
Object
?
Image
25
Images

• Virtual Image - no light reaches image
• Real Image - light rays reach image
• Erect or Upright Image directions unchanged
  (object up is same direction as image up)
• Inverted Image directions opposite (object up
  is image down)
• Reversed Image left seems right

26
Spherical Mirrors

• Parameters of a spherical mirror
• Principal axis line through C and V where V is
  the center of the mirror
• C Center of Curvature center of sphere or
  circle (2 dimensions)
• R Radius of Curvature
• F Focal point
• Radius lines are always
• normal to surface
• of circles and spheres

V
R
C
principal axis
F
Concave
27

• Given an object to the left of C, by the law of
  reflection, ? ?, light rays will converge at a
  point between V and C for rays with small angles
  to the principal axis
• For larger angles the rays will intersect at
  different points creating spherical aberration
  See Fig. 36.8

V
C
principal axis
O
I
Concave
28
Image Concave Mirror

• Rays through C are normal to surface ?0
• Draw rays to V (intersection of surface and
  principal axis) ? ? and through C, ?0
• Tan ? h/p -h/q, since M h/h, -q/p
• Note -h inverted
• Image is
• smaller
• inverted
• real
• outside point F

Concave
p
R
h
V
?
a
a
h
C
?
F
q
29

• since tan a -h/(R-q) h/(p-R)
• h/h -(R-q)/(p-R) eqn 36.3
• 1/p 1/q 2/R eqn 36.4 MIRROR EQN
• For a very distant object where p-gt infinity, 1/p
  0 so 1/q 2/R
• This case it is called the focal point F, where F
  2/R eqn 36.5
• hence 1/p 1/q 1/f, eqn 36.6, the mirror
  equation
• The focal point is where rays parallel to the
  axis pass through

30
Spherical Convex Mirror

• Often called a diverging mirror
• Concepts presented are valid for this type of
  mirror as well if adhere to the following
  procedure
• R, F q negative
• p, h h positive

R
C
V
principal axis
F
p
q
31
Procedure

• Front side of mirror where light waves
  originate and move towards the mirror
• Back side is the other side

32
Mirror Sign ConventionsTable 36.1
33
Ray Trace Example 1

• Object outside focal point
• Image inverted and smaller
• Rays drawn through C, and parallel to principal
  axis and through F

V
C
F
34
Ray Trace Example 2

• Object is inside the focal point F
• Image is virtual, upright and magnified

V
C
F
35
Ray Trace Example 3

• Convex mirror
• Image is virtual, reduced and upright

R
C
V
F
p
q
36
Ray Tracing

• Principal axis goes through C, center of
  curvature so is perpendicular to the surface at
  V. Bottom of object.
• Ray 1 top of object parallel to principal axis
  goes through Focal point or reflects away from
  focal point for convex mirror
• Ray 2 top of object through focal point
  reflects parallel to axis where it intersects
  mirror
• Ray 3 top of object through Center of curvature,
  reflects back on self

37
Example Problem 1

• Given object of height 1 cm located 30cm in front
  of a concave mirror of focal length 10 cm, what
  is a) radius of curvature? b) Location of image?
  c) Real/virtual image? d) Magnification? e)
  Enlarged? f) Image height? g) upright/inverted?

38
V
C

• h 1cm, p 30cm, f10cm
• f R/2, 1/p 1/q 1/f
• M -h/h, M -q/p
• R 2f 20cm
• q 1/(1/f 1/p) 15cm
• Real
• M -q/p -15/30 -0.5
• Reduced
• M-h/h , -0.5 -0.5cm/1cm
• Inverted

F
39
Chapter 36 Refractive Surfaces and Lenses
PHYS 2326-32
40
Concepts to Know

• Spherical Refracting Surfaces
• Thin Lenses

41
Definitions

• Plane surface one that is flat
• Concave surface, one that is curved inward at the
  center
• Convex surface, one that is curved outward

Eye
Plane
Concave
Convex
42
Differences From Mirror

• Light goes through
• There are effects from the index of refraction
• Sign Conventions must change

43
Plane Surface Boundary
p
q

• For a plane mirror
• pq and M 1
• What about for a fish tank?

h
?
Object
?
Image
44
Plane Refraction

• Unlike the mirror
• viewing into another
• index of refraction
• medium is affected
• even by a flat plane

q
n1.30
n1.00
Object
p
45
Refraction

• Fig. 36.17 shows a spherical boundary between two
  media of different indices of refraction
• Using Snells law and the small angle
  approximation one can achieve eqn 36.8

?2
?1
n2
n1
?1
d
a
?
ß
I
C
O
R
p
q
46
Refracting Surface

• Eqn 36.8 provides the result of the geometric
  derivation from Fig. 36.17
• Notice that the image distance depends upon the
  object distance, Radius of curvature and the
  index of refraction for each media
• Note too, that if R becomes infinite we have a
  flat plane surface and (n2-n1)/R becomes 0 so
• q -(n2/n1)p eqn 36.9

47
Lenses

• Lenses have 2 surfaces, not just one
• A lens may have a concave, convex, or plane
  surface referred to as plano when being
  described
• A lens is converging or diverging
• Converging lens types include biconvex,
  concave-convex, and plano-convex
• Diverging lens types include biconcave,
  convex-concave, and plano-concave

48
Refracting Surface Sign Conventions
49
Lens

• Eqn 36.10 starts with eqn 36.8 and assumes n1
  1.00 and n2 n for surface 1
• Eqn 36.11 is for surface 2 and differs because
  the light is going from inside the material out
  to the air, n1 n and n2 1.00 for surface 2
• The image formed by surface 1 becomes the object
  for surface 2

50
Thin Lens Equation

• Note that p2 -q1 t and p2 will be positive
  by the sign convention because q1 has a negative
  value
• if t is very small, p2 -q1 (for both real or
  virtual image)

R1
R2
n1
I
O
t
C1
p
q1
p2
51
Thin Lenses

• Eqn 36.11 becomes Eqn 36.15, The Lens Makers
  Equation

52
Lens Makers Equation

• Can be used to determine R1 and R2 for a lens
  given a desired focal length and an index of
  refraction
• If given the index of refraction and radii of
  curvature (R1 and R2), one can find the focal
  length
• If assume n is the ratio of the index of
  refraction between the lens and what it is
  immersed in other than air the eqn can be used

53
Image Magnification

• M h/h -q/p which is the same as for
  mirrors.
• When M is positive, image is upright and on the
  same side of the lens as the objec
• When M is negative, the image is inverted and on
  the opposite side of the lens

54
Thin Lens Equation

• Eqn 36.15, the lens makers equation can be used
  with 36.14 to create 36.16, a thin lens equation
  exactly like that of mirrors.
• Note that focal points are the same distance on
  both sides of the lens
• Also, a converging lens is thicker in the middle
  while a diverging lens is thinner in the middle

55
Table 36.3
56
Ray Tracing for Converging Lens

• Ray 1 parallel to principal axis refracted
  through focal point on back side of lens
• Ray 2 draw through lens center continues straight
  line
• Ray 3 through focal point on front side or as if
  coming from focal point (if inside f) and
  emerging parallel to principal axis

57
Ray Tracing for Diverging Lens

• Ray 1 drawn parallel to principal axis, emerges
  away from the focal point on the front side
• Ray 2 drawn through lens center in straight line
• Ray 3 drawn towards backside focal point emerges
  parallel

58
Example Problem 1

• A thin double concave lens has a radius of 10cm
  on both sides and made from glass with an index
  of refraction 1.5. A light bulb 2cm tall is
  located 50cm in front of the lens along the
  principal axis. a) What is the focal length of
  the lens? Converging or diverging? b) Where is
  the bulb image? c) What is the image height?
  Enlarged or reduced? Erect or inverted?

59

• Equations
• 1/f (n-1)(1/R1 1/R2)
• 1/p 1/q 1/f
• M -q/p h/h
• a) f? 1/f (1.5-1)(1/(-10) 1/(10))-0.5(2/10)
• f -10cm (diverging)
• b) 1/50cm 1/q 1/(-10), q -12.5cm
• on front side and it is virtual
• c) m -12.5/50 0.5cm reduced, erect

60
Problem 36.33

• 36.33 An object located at 20cm to left of a
  diverging lens with focal length f -32.0cm.
  Determine a) the location and b) the
  magnification of the image. c) Construct a ray
  diagram

61

• The problem provides p object distance and f
  focal length
• Can use thin lens eqn to solve for q image
  distance
• Can then use magnification definition to
  determine the magnification
• Ray tracing for a diverging lens

62
Ray Tracing for Diverging Lens

• Ray 1 drawn parallel to principal axis, emerges
  away from the focal point on the front side
• Ray 2 drawn through lens center in straight line
• Ray 3 drawn towards backside focal point emerges
  parallel

63
Chapter 36 Optical Instruments
PHYS 2326-33
64
Concepts to Know

• f-number
• Fresnel Lens
• Eye
• Magnifier
• Microscope
• Telescope
• Aberrations

65
Combination of Thin Lenses

• Image formed by first lens must be located as if
  the second lens is not present
• A ray diagram is drawn for the second lens using
  the image of the first lens as the object of the
  second
• The second image is the image of the system
• If the first image is behind the second lens, it
  is virtual and has p2 for its location
• Magnification is a product of the individual
  magnifications M M1M2

66
Two Thin Lenses in Contact

• Given 2 thin lenses in contact with each other of
  focal lengths f1 and f2, this compound lens has a
  focal length
• 1/f 1/f1 1/f2 Eqn 36.19

67
Lens Aberrations

• Our equations make assumptions like small angles
  to the axis, that lenses are thin and even that
  the index of refraction is uniform at all
  wavelengths.
• Ray tracing permits precise analysis using
  Snells law.
• There are two common abberrations
• spherical aberration
• chromatic aberration

68
Spherical Aberration

• Caused by parallel rays coming in further away
  from the principal axis and intersect the
  principal axis after refraction at different
  points than rays closer in
• Parabolic mirrors and lenses are better but are
  much harder to make
• Parabolic mirrors are common in telescopes
• Cameras use adjustable apertures to control the
  amount of light and to reduce spherical aberration

69
Chromatic Aberration

• Chromatic aberration is related to dispersion and
  is caused by differences in the index of
  refraction for different wavelength. Shorter
  wavelengths like violet may refract more than
  red, changing the focal point inward
• Solutions to this are the use of more exotic
  lenses such as achromatic or apochromatic which
  are composed of two or even three different
  materials, some are more exotic than plain glass.

70
Cameras Eyes

• Cameras and eyes are quite similar
• Both have a lens and an aperture (usually
  adjustable to control incoming light)
• Both have a light sensitive medium at the back
  which captures the image
• Eyes tend to have a spherical sensing area which
  permits superior results with simpler optics
• Cameras tend to have a flat sensing area
• Both must create a real image on the sensor

71
Aperture

• Some optical systems, like most cameras have an
  adjustable aperture
• All optical systems have an aperture
• Aperture is f/D where f is the focal length and D
  is the diameter
• Apertures are called f-numbers
• Aperture is related to the amount of light that
  is permitted to enter which is proportional to D2

72

• f-number is defined by f/D Eqn 36.20
• Intensity I is proportional to 1/f-number2
• A telescope usually doesnt have an adjustable
  aperture but rather D is the diameter of the main
  lens or the main mirror.
• The lower the f/number, the faster the lens and
  often the more expensive it is for a certain
  quality grade

73
Simple Magnifier

• Even with a magnifying glass, the size of the
  object viewed by the eye depends upon the angle ?
  subtended by the object
• A converging lens (or magnifying glass) may be
  used to increase the apparent size of an object
  as seen by the eye.
• This is an angular magnification m ?/?o
• Assuming ones vision permits 25cm as the closest
  distance, eqn 36.24 mmax 1 25cm/f is the
  maximum magnification possible and mmin 25cm/f
  is the lowest for a magnifier

74
Compound Microscope

• A compound microscope consists of an objective
  lens near the sample being observed an an
  eyepiece
• The objective has a very short focal length lt 1cm
  and the eyepiece with f a few cm
• These lenses are placed at a distance L greater
  than the focal lengths of the lenses
• The object is placed just outside the focal point
  of the objective

75
Microscope Magnification

• Overall magnification is M Mome where Mo is the
  lateral magnification of the objective and me is
  the angular magnification of the eyepiece
• M -L/fo (25/fe) the sign indicates an
  inversion

76
Telescope

• Eqn 36.27 is shown in section 36.10 along with
  the approximations and derivation
• m ?/?o - fo/fe ratio of the focal length of
  the objective divided by that of the eyepeice
• The two common types are reflectors and
  refractors with mirrors or lenses for the
  objective
• The largest reflectors are 10 meters in diameter
  while the largest refractor is about 1meter.

77
Chapter 37 Interference
PHYS 2326-34
78
Concepts to Know

• Interference
• Principle of Superposition
• Monochromatic Light
• Coherent Light
• Antinodal Curves (Constructive Interference)
• Nodal Curves (Destructive Interference)
• Interference Fringes
• Phase Shift (Reflection off Slow Medium)
• Thin Films

79
Inteference

• Like sound waves, light waves can produce
  interference
• Nodes exist where there are minima (dark areas)
  from destructive interference
• Antinodes exist where there are bright areas from
  constructive interference

80
Monochromatic Light

• Monochromatic light simply means one color
  which is 1 frequency or wavelength
• Lasers produce monochromatic light very well
• Sodium vapor lights do a fair job although they
  are producing bichromatic light but there are a
  variety of arc lamps which are monochromatic
• Lightbulbs produce a wide range of wavelengths

81
Coherent Light

• Coherent light refers to light having a
  consistent phase between sources.
• From our study of sound, we determined that
  phasing between PA system speakers could have
  negative effects on ones hearing at some
  locations
• In order to see interference effects we must have
  consistent phase.

82
Conditions for Interference

• Sources must be coherent have a constant phase
• Sources must be monochromatic, a single wavelength

83
Youngs Double-Slit Experiment

• Performed by Thomas Young in 1800
• It showed Huygenss principle of subdividing wave
  fronts into new wavelets or wave fronts to be
  correct

84

• S1 and S2 are narrow slits separated by distance
  d. Distance d is the difference in path lengths
  r1 and r2 and ? is the angle from the normal at
  the slit center line and y is the offset distance
  at the screen
• When d is a multiple of wavelength there is
  constructive interference

r1
y
S1
r2
?
d
S2
d
L
85

• d sin ?bright m?
• d sin ?dark (m ½) ?
• since tan ? y/L
• ybright L tan ?bright
• ydark L tan ?dark
• for small angles where ? sin ?
• ybright L (m ? /d)

86
Phase Shift from Reflection

• Section 37.5 refers to Lloyds mirror where our
  two sources are one source plus a reflected image
  of the source
• When an electromagnetic wave is reflected from
  the interface to a medium with a higher index of
  refraction than the one in which it is traveling,
  it undergoes a 180 degree phase reversal.

87

• When a reflection occurs at a surface where the
  index of refraction is less than that of the
  medium the wave is traveling in, there is NO
  phase reversal at the reflection
• This corresponds to a ½ wavelength displacement
  shift or a ½ integer shift

88
Thin Films

• Thin films such as oil on water or soap bubble
  surfaces exhibit interference effects by showing
  varied colors when white light is incident on the
  films.
• This occurs because the reflection from the top
  surface has a 180 degree phase shift and the
  second surface at some distance t does not have
  this phase shift and the total effective path
  difference becomes a multiple of certain
  wavelengths
• 2t (m1/2)?n where ?n ? /n (index of
  refraction)
• 2nt (m1/2) ? for constructive interference and
• 2nt m ? for destructive interference (eqn
  37.15-17)

89
Color Wavelength
500nm
400nm
600nm
700nm
blue
green
red
yellow
cyan
magenta

• Primary colors are red green and blue and have a
  range of wavelengths for each. Adding any
  combination creates other colors. Often called
  RGB
• Cyan magenta and yellow are combinations of two
  primary colors
• Wavelengths are shown in nanometers (10-9 meters)
  sometimes these are shown in Angstroms which is
  10-10

90
Example Problem 1

• In a double slit interference experiment, the
  slits are 10 micron (10-6 meters) apart and the
  screen is 2 meters away. If 500nm wavelength
  light is used, find a) the location of the first
  dark fringe, b) the location of the 3rd bright
  fringe, c) the spacing between fringes, d) the
  theoretical number of bright fringes possible.

91

• a) d sin ? (m1/2)?, 1E-5 sin ? (1/2) 5E-7 ?
  1.43
• y L tan ? 2.0 tan(1.43) 0.05m
• b) d sin ? (m)?, 1E-5 sin ? (3) 5E-7
• ? 8.63
• y L tan ? 2.0 tan(8.63) 0.30m
• c) d sin ? (1)?, ? 2.86
• y 0.10m
• d) let maximum ? 90, d sin 90 (m)?, m20.
  This is for 1 side and there is a middle fringe
  total 41

92
Example Problem 2

• What is the minimum thickness of a soap bubble
  film with index of refraction 1.33 that would
  reflect 650nm most brightly? b) What is the
  minimum thickness for an anti-reflecting coating
  of index of refraction 1.4 or a glass of index
  1.5 which would reflect no green light of
  wavelength 550nm? c) what would be the color of
  the light that is reflected off the lens

93

• n 1.33, ?o 650nm, ? ?o /n 488nm, ?m 0
  m2-m1
• m2 2d/ ?, since m1 ½ (due to 180 deg.
  inversion not present at m2) our path difference
  can be ½ wavelength. Get wavelength inside
  material and determine d m2 ?/2 ½ 488 /
  2122nm
• b) coating n1.4, glass 1.5 at 550nm ? ?o /n
  393nm, m1 ½, m2 2d/ ? ½ since both m1 and
  m2 reflect from greater index of refraction
  mediums ?m ½ m2-m1 m2 ½, m2 1
• ? 550/1.4 393,
• hence m2 1 2d/ ? ½, 2d/ ? 1/2, d ?/4
  393/4 98nm
• c) green transmitted, blue red reflected,
  magenta

94
Chapter 38 Diffraction Polarization
PHYS 2326-35
95
Concepts to Know

• Diffraction
• Single Slit Diffraction
• Multiple Slit Diffraction
• Diffraction Grating
• Mth order Lines
• Grating Spectrometer
• Resolution or resolving power
• Circular Aperture Diffraction
• Polarization

96
Narrow Slits

• Like the dual slit, a narrow slit produces a
  pattern because each portion of the slit acts as
  a source of light waves

5
4
a/2
y
3
?
a
2
a/2
1
(a/2)sin?
D
97

• If dark fringes occur at a sin ? m? then
  where do light fringes occur?
• hint its halfway in between the dark fringes
• By geometry tan ? y/D which is valid for large
  angles as well as small angles
• Using small angle approximation and substituting
  for tan ? sin ?,
• y Dm ?/a

98
Resolution

• Rayleigh criterion is when the central maximum of
  one image falls on the first minimum of a second
  image, the images are said to be just resolved
• for a slit of width a, this is ?min ?/a
  assuming wavelength ltlt a, eqn 38.5
• for a circular aperture ?min 1.22 ?/D where D
  is the diameter

99
Example

• for 600nm red light, our TAMUK 16 telescope
  (400mm 0.4m) has a resolving power ?min 1.22
  ?/D (1.22)(6.0E-7)/0.4 0.00000183 radians
  0.38 arcseconds
• since 2pi radians 360 deg, 1 deg60 arcminutes
  and 1 arcminute 60 arcseconds so radians 360
  60 60/(2 pi) gives arc seconds
• Excellent viewing is essentially never better
  than 0.5 arcseconds due to atmospheric
  turbulence, this doesnt come into play for
  visual work. To get higher resolution, larger
  telescopes use adaptive optics to correct for
  turbulence in real time. Amateurs sometimes
  either use crude imitations or take lots of short
  exposures for computer enhancement processing to
  achieve better results than viewing would permit.

100
Diffraction Grating

• Diffraction gratings are like multiple slits,
  usually with thousands of lines (crude slits) per
  inch, if not tens of thousands of lines
• It may be a reflective or a transmission grating
• Variable d usually refers to the spacing between
  the lines and one is interested in the positions
  of the maxima

101

• d sin ?bright m? where m0,/- 1,2,3
• eqn 38.7
• m 0 is the 0th order maximum which doesnt
  separate out by wavelength as ? 0
• m 1 for the first order wavelengths, 2 for the
  second and note that when m is not 0, ?bright
  depends upon wavelength
• Note too figure 37.8 shows that the number of
  slits used increases the sharpness

102

• Also, note that when the wavelength doubles, like
  from violet to red, the angle for the second
  order violet may be greater than the angle for
  the first order red so there could be overlap

103
Diffraction Grating Applications

• Gratings provide an excellent way to break up
  light into different wavelengths and provide a
  linear result unlike a prism that depends upon
  the change of the index of refraction with
  wavelength in order to work

104
X-ray Diffraction

• While diffraction works well with visible light,
  it turns out that with shorter wavelengths, can
  be used to study much smaller items. X-rays have
  been used to study the structure of crystals with
  a spacing between molecules of around 0.1nm
• 2d sin ? m? where m1,2,3 describes the
  condition for constructive interference known as
  Braggs law, eqn 38.8 for reflections from atomic
  planes in a crystal

105
Polarization

• Light, like radio waves, is referred to as
  electromagnetic radiation because it has an
  electric field E and a magnetic field B which are
  perpendicular to each other and both are
  perpendicular to the direction of travel
• Light normally consists of a bunch of waves in
  superposition that are randomly oriented
  according to their electric field E.
• Polarization refers to the direction of the
  electric field vector E.

106

• Light, like radio waves, can be polarized or
  unpolarized and there are a variety of
  polarizations possible
• Light from an incandescent light bulb is
  unpolarized
• Radio waves from a vertical antenna are said to
  be vertically polarized as the electric field
  varies in the vertical direction. These are also
  linearly polarized as the electric field is in
  one direction that doesnt change with time

107

• Given two vectors, E and c, the electric field
  and the wave velocity or direction of
  propagation, a plane is formed and the wave is
  said to be plane polarized
• One may create a plane polarized wave by either
  creating it that way as in radio or by removing
  or filtering out all other polarizations
• There are 4 ways to create a linearly polarized
  beam from an unpolarized beam

108
Selective Absorption

• Polaroid a film that selectively filters out
  light through absorption, Invented by Land in
  1938
• Commonly used in sun glasses
• Composed of 2 sheets, the polarizer and the
  analyzer rotated by an angle ?

109
Maluss Law

• Intensity of the transmitted polarized beam from
  the polarizer to the analyzer is Imax and the
  output from the analyzer intensity is I Imax
  cos2 ? eqn 38.9 and is known as Maluss law and
  applies to two polarizing materials whose
  transmission axes are at an angle ? to each other
• I is maximum when ? 0 or 180 and
• I 0 when ? 90

110
Polarization by Reflection

• Polaroid sun glasses are popular because
  reflected light tends to become polarized
• Even though light may be unpolarized, when it is
  reflected it may become completely or partially
  polarized
• Reflection has a preference for waves with
  electric fields parallel to the surface

111

• If ?1 is varied so that the angle between the
  refracted ray and the reflected ray 90 the
  reflected beam is totally polarized while the
  refracted beam is partially polarized. This
  angle is the polarizing angle ?p and from Snells
  law n2/n1 sin ?p /sin ? and by geometry tan ?p
  n2/n1 eqn 38.10 Brewsters law

?1
?1
?1
?1
n1
n1
n2
90
n2
?2
?2
112
Polarization by Double Refraction

• Some materials like calcite and quartz are double
  refracting or birefringent materials which can
  split an unpolarized beam into an ordinary O ray
  and an extraordinary E ray which are mutually
  perpendicular in polarization and have different
  indices of refraction
• There is an optical axis where the indices of
  refraction are the same
• Some materials exhibit this when stressed and
  optical stress analysis may be used in
  mechanical design work

113
Polarization by Scattering

• Why is the daytime sky blue?
• Its called scattering and occurs from
  particulates and primarily from molecules of
  oxygen and nitrogen the most predominant
  molecules in the atmosphere
• Because the molecular size is around 0.2nm and
  much smaller than visible wavelengths and the
  scattering varies by 1/?4, shorter wavelengths
  such is violet and ble are scattered more
  efficiently and hence has higher intensity than
  longer wavelengths such as red.
• Since eyes are less sensitive to violet, color
  shows up more as blue even though theres less
  intensity

114
Nonlinear Polarization

• Aside from linear polarization there are other
  forms of polarization
• Circular polarization
• Elliptical polarization
• Helical antennas are circularly polarized and
  work better transmitting through the atmosphere
  to space
• Light may also be circularly polarized when
  composed of two waves 90 degrees apart in
  polarization and out of phase by 90 degrees

115
Example

• Given a single slit with width of 0.2mm and 600nm
  laser light casting a pattern on a screen 6m
  away, a) what is the linear spacing between two
  sucessive fringes on the screen? b) how far from
  the central bright fringe is the 4th dark fringe?
  c) How far from the central (zeroth) bright
  fringe is the first bright fringe located?

116

• a 2.0E-4m, ? 6.0E-7m D6m, a sin ? m ?, y
  Dm?/a
• a) let m1 for distance from m0, y 0.018m
• b)m4, y Dm?/a 0.072m
• c) bright fringes are ½ values, first dark fringe
  is m1 so first light is m 1.5
• yDm?/a 6 1.5 6.0E-7 / 2.0E-4 0.027m

117
Example 2

• Diffraction grating with 500 lines per mm. a)
  What is the angle between red light of wavelength
  700nm and violet light at 400nm for the second
  order spectrum?
• b) Does the 3rd order spectrum overlap the
  second? c) If a beam of green light of wavelength
  550nm passes through the grating and is projected
  onto a wall 4 m away, what is the distance
  between the first order fringe and the second
  order fringe?

118

• d 0.001mm/500 2.0 E-6m or 2.0 µm
• d sin ? m?
• red 2.0E-6 sin ? 2(7.0E-7), ? 44.4
• violet 2.0E-6 sin ? 2(4.0E-7), ? 23.6
• ?? 44.4 23.6 20.8
• b) 3rd order, m3, d sin ? 3?
• 2E-6 sin ? 3 (4.0E-7), ? 36.9 smallest angle
  of 3rd order and largest for 2nd order is red
  44.4, they overlap significantly

119

• c) distance at 4m between 1st and 2nd order
  fringes for green light 550nm
• d sin ? m?, for m1, 2E-6 sin ? (1)(5.5E-7),
  ? 16.0, for m2, ? 33.4
• for m1, y D tan ? 4 tan(16) 1.14m
• for m2, y 2.63m
• ?y 2.63-1.14 1.49 m