Chs 37
Huygen’s Principleand 38
•Any wave (including electromagnetic waves) is able to propagate
because the wave here affects nearby points there
•In a sense, the wave is the source for more of the wave
•A wave here creates waves in all the forward directions
•For a plane wave, the generated waves add up to make more plane
waves
•Mathematically, this works, but for plane
waves, no one does it this way
Diffraction Through a Tiny Hole
•The waves come out in all directions
•It is only because the whole wave makes new waves that the
waves add up to only go forwards
•What if we let the wave pass through a tiny hole?
•Smaller than a wavelength
•Only one point acts as source
•Waves spread out in all directions
r
E  E0 sin  kx  t 
Diffraction is bending
E0
E
sin  kr  t 
r
•What’s interesting is that
oscillations depend on distance
from slit
E
sin  kr  t 
CT -1 - Diffraction occurs when light passes a:
A. pinhole.
B. narrow slit.
C. wide slit.
D. sharp edge.
E. all of the above
Ans E
Interference Through Two Slits
•Now imagine we have two slits, equally sized
•Each slit creates its own waves
•In some directions, crests add with
crests to make bigger “brighter”
crests
•In others, crests combine with
troughs to make minimum areas
•In the end, what you get is a
pattern of alternating light and dark
bands
•We’re about to need an obscure math identity:
 A B 
 A B 
sin A  sin B  2 sin 
 cos 

2
2




Two Slit Interference
•What do the EM waves look like far away?
•Let the separation of the slits be d
•Let’s find total E-field at point P
d sin
E  E1  E2~ sin  kr1  t   sin  kr2  t 
 2 sin  krave  t  cos  12 k  r2  r1  
d

r2
rave 
1
2
 r1  r2 
r2  r1  d sin 
r1
E
2sin  krave  t  cos  12 kd sin  
I
E2
sin 2  krave  t  cos2  12 kd sin  
P
Two Slit Interference (2)
I  cos
2

1
2
kd sin  
k
2

•Where is it bright?
•Where is it dark? sin  bright  m d
  d sin  
I  I max cos 



2
m  0, 1, 2,
sin dark   m  12   d
CT -2- An interference pattern is formed on a screen by shining a planar wave on a
double- slit arrangement (left). If we cover one slit with a glass plate (right), the
phases of the two emerging waves will be different be-cause the wavelength is
shorter in glass than in air. If the phase difference is 180°, how is the interference
pattern, shown left, altered?
A.
B.
C.
D.
E.
The pattern vanishes.
The bright spots lie closer together.
The bright spots are farther apart.
There are no changes.
Bright and dark spots are interchanged.
Ans E
Ex- (Serway 37-10) In a location where the speed of sound is 354 m/s, a 2000 Hz
sound wave impinges on two slits 30.0 cm apart. (a) At what angle is the first
maximum located? (b) If the sound wave is replaced by 3.00-cm microwaves,
what slit separation gives the same angle for the first maximum? (c) If the slit
separation is 1 M, what frequency light gives the same first maximum angle?
Solve on
Board
Phases
•When you combine two (or more) waves, you need to know the phase
shift between them:
E  A sin  x   B sin  x   
•The angle is the phase shift
•When the phase shift is zero, the waves add constructively
•The result is bigger
•Same thing for any even multiple of 
•When the phase shift is , the waves add destructively
•The result is smaller
•Same thing for any odd multiple of 
•To find maximum/minimum effects, set phase shift to even/odd
multiples of 
Phase Shift From Traveling
•As a wave passes through any material, its phase shifts
•For a distance d, we have:
  kd

E  sin  kx  t 
2 d

•Recall, wavelength  changes inside a material
c
f 
n
f 0  c

0
n

2 dn
0
Light of wavelength 0.5 m takes two paths, both of length 1 m,
one through air, the other through glass (n = 1.5). What is the
difference in phase between the two waves in the end?
A) 0 B)  C) 2 D) 3 E) None of the above
1 m
1  2 11.5 0.5  6
2  2 11.0 0.5  4
  2
Ex- (Serway 37-25) The intensity on the screen at a certain point in a double slit
interference pattern is 64% of the maximum value. (a) What minimum phase
difference (in radians) between the sources produces this result? (b) Express this
phase difference in terms of a path difference for 486.1 nm light.
Solve on board
Four Slit Interference
•What if we have more than two slits?
•Four slits, each spaced distance d apart
•Treat it as two double slits
E  E1  E2  E3  E4  E1,2  E3,4
 2sin  kr1,2  t  cos  12 kd sin  
r3,4
2sin  kr3,4  t  cos  12 kd sin  
 4 sin  krave  t  cos  d sin   
r1,2
P
•For four slits, every
third band is bright
 cos  2 d sin   
I  E2
More Slits and Diffraction gratings
•This process can be continued for more slits
•For N slits, every N – 1’th band is bright
•For large N, bands become very narrow
•A device called a diffraction
N=8
grating is just transparent with
N = 16
closely spaced regular lines on it
N = 32
•You already used it in lab
sin  bright  m d
m  0, 1, 2,
•Diffraction gratings are another way to divide light into different colors
•More accurate way of measuring wavelength than a prism
•Commonly used by scientists
CT -3 - A diffraction grating is illuminated with yellow light at normal incidence.
The pattern seen on a screen behind the grating consists of three yellow spots, one
at zero degrees (straight through) and one each at ±45°. You now add red light of
equal intensity, coming in the same direction as the yellow light. The new pattern
consists of
A.
B.
C.
D.
E.
F.
red spots at 0° and ±45°.
yellow spots at 0° and ±45°.
orange spots at 0° and ±45°.
an orange spot at 0°, yellow spots at ±45°,
and red spots slightly farther out.
an orange spot at 0°, yellow spots at ±45°,and red spots slightly closer in.
Ans D
Resolution of Diffraction Gratings
•Note that the angle depends on the wavelength
•With a finite number of slits, nearby wavelengths
may overlap
N=8

1.1
sin 1  m1 d
sin  2  m2 d
•The width of the peaks is about
  sin     dN
•The difference between peaks is
  sin    m    d
•This quantity (mN) is called the
resolving power
•Even if N is very large,
effectively N is how many slits
the light beam actually falls on
•We can distinguish two peaks if:
m    d   Nd

mN 

Diffraction through a single slit
•What if our slit is NOT small compared to a wavelength?
•Treat it as a large number of closely spaced sources, by Huygen’s
principle
•Let the slit size be a, and r the distance to the center
ave
•Let x be the distance of some point from the center
•The distance r will be slightly different from here to P
a x
r
rave
E
1a
2
 12 a
r  rave  x sin 
E ~ sin  kr  t 
 sin  krave  kx sin   t 
sin  krave  kx sin   t  dx
a/2
1

cos  krave  kx sin   t    a / 2
k sin 
P
Diffraction through a single slit (2)
1


cos
kr


1
ave
2 ka sin   t 
E



k sin    cos  krave  12 ka sin   t  
2
1
1
sin  2 ka sin  
 sin  2 ka sin   

sin  krave  t  I  I max 

1
1
k
sin

ka
sin

2
 2

I  I max
 sin  a sin    



a
sin




2
sin dark  m a
m  1, 2,
•Very similar to equation for multi-slit
diffraction, but . . .
•a is the size of the slit
•This equation is for dark, not light
•Note m= 0 is missing
•Central peak twice as wide
Ex - (Serway 38-7) A screen is placed 50 cm from a single slit, which is
illuminated with 690 light. If the distance between the first and third
minima in the diffraction pattern is 3 mm, what is the width of the slit?
Solve on board
If you used a little wider slit,
the pattern would
A) Get wider and dimmer
B) Get wider and brighter
C) Get narrower and dimmer
D) Get narrower and brighter
Screens and Small Angles
•Usually your slit size/separation is large compared to the wavelength
•Multi-slit: sin 
 m d Diffraction: sin 
 m a
bright
dark
•When you project them onto a screen, you need to calculate locations of
these bright/dark lines
x
•For small angles, sin and tan are the same L  tan   sin 
x
L
xbright
L
m
d
xdark
L
m
a
Diffraction and Interference Together
a
•Now go through two finite sized slits
•Result is simply sum of each slit
•Resulting amplitude looks like:
d
a
E
sin  12 ka sin   sin  kr1  t  


1
  sin  kr2  t  
2 k sin 
 sin  a sin    
2   d sin  

 sin 




  a sin   
2
I  I max
a = d/5
•Resulting pattern has two kinds of
variations:
•Fast fluctuations from separation d
•Slow fluctuations from slit size a
Diffraction Limit:
•When light goes through a “small” slit, its
direction gets changed
•Can’t determine direction better than this
sin min 
min
 a
 a
•If we put light through rectangular (square) hole,
we get diffraction in both dimensions
•A circular hole of diameter D is a trifle smaller,
which causes a bit more spread in the outgoing wave
•For homework, use this formula; for tests, the
approximate formula is good enough
min  1.22  D
min
 D
a
a
D
CT -4 - For a given lens diameter, which light gives the best resolution in a
microscope?
A. red
B. yellow
C. green
D. blue
E. All give the same resolution.
Ans D
Diffraction Limit (2)
If the pupil of your eye in good light is 2 mm in diameter, what’s
the smallest angle you can see using 500 nm visible light?
min
1.22  5 107 m
4
 1.22 D 

3.05

10
rad
3
2 10 m
 1 arc-min
•A degree is 1/360 of a circle, an arc-minute is 1/60 of
a degree, an arc-second is 1/60 of an arc minute
•Telescopes require large apertures to see small angles
Reflection and Phase Shift
•When you reflect off of a mirror, the reflected wave must cancel the
incoming wave
•It has a  phase shift
•When you go from a low index of
refraction medium to a high one,
some of the wave is reflected
•It also has a  phase shift
 phase
shift
 phase shift
•When you go from a high index
of refraction medium to a low
one, some of the wave is reflected
•This has a 0 phase shift
0 phase shift
Concept Question
Suppose we are in a glass medium, and we have a wave that goes
from glass to air to glass. If the layer of air is much smaller than
one wavelength, then the two reflected waves will add
A) Constructively
B) Destructively
C) Insufficient Info
•First transition: high to low
•no phase shift
•Second transition: low to high
• phase shift
•Compared to each other, the two waves are  out of phase with
each other
•They will have a tendency to cancel
•Very little effect from layer if thinner than a wavelength
Interference from Thin Films
•Suppose we go through a thin soap film
•Index goes up then down
Front surface:
•Phase shift of  from reflection (low-high)
t
Back surface:
•Phase shift of 2t/ from traveling
•Phase shift of 0 from reflection
•Phase shift of 2t/ from traveling
4 t


Total phase shift between two reflected waves:

•Weak reflection when odd times :
4 tweak
•Strong reflection when even
  2m  1 

•Same results for index down then up

•Opposite for:
2tweak   m
•Index up, then up
•Index down, then down
1
2tstrong    m  2 
Applications of Thin Film Interference
•What if the light isn’t monochromatic?
2tweak  m
2tstrong    m  12 
•Some wavelengths are enhanced, others are not
•Soap bubbles
•Oil on water
•Newton’s rings: convex lens on flat glass plate
•Air gap changes thickness in circular pattern
•Alternating light/dark regions
d
narrow
air gap
Ex- (Serway 37-34) An oil film (n = 1.45) floating on water is illuminated by white
light at normal incidence. The film is 280 nm thick. Find (a) The dominant
observed color in the reflected light and (b) the dominant color in the transmitted
light.
Solve on Board
Michelson Interferometer
•Interference easy to measure
•Can see much smaller than one wavelength
•LIGO, state of the art, can see 10-15 m!
Laser
Detector
Hanford, Washington
Mirrors
Crystal Scattering of X-rays
•Mysterious rays were discovered by Röntgen in 1895
•Suspected to be short-wavelength EM waves
•Order 1-0.1 nm wavelength
•Scattered very weakly off of atoms
•Bragg, 1912, measured wavelength accurately
2d cos  m
 
d
•Scattering strong only
if waves are in phase
•Must be integer
multiple of wavelength
Polarization
•Recall that light waves have electric and magnetic fields
perpendicular to the direction of motion
•But there are two independent ways of arranging this
•Called polarization
•Our eyes can’t tell these two polarizations apart
•But some instruments can measure or take advantage of
polarization
•We describe polarization by telling which direction the
electric field points, e.g. vertically or horizontally
•A polarizer polarizes light along its transmission axis.
•Malus’s Law
2
I = I0 cos (θ)
B0
E0
E0
B0
Methods of Producing Polarization (1)
Direct production
•Antennas produce waves that are automatically polarized
Scattering
•Light waves of all orientations hit small targets
•Target has vibrating charges, like an antenna
Reflection and Brewster’s Angle:
•When light hits a substance, some of it reflects and some refracts
•Fraction of each depends on polarization
•There’s a special angle – Brewster’s angle – where reflected is
completely polarized
tan   n n
P
n1
n2
P
E0
2
1
+
+
+
+
+
+
–
–
–
–
–
–
Methods of Producing Polarization (2)
Birefringent Crystals
•Index of refraction has to do with electric fields from the wave pushing
atoms around
•In some crystals, it is easier to push them one way than another
•Index of refraction depends on polarization
•You can use such birefringent crystals to sort light based on polarization
Selective absorption
•Similarly, some materials absorb one polarization better than another
E0
E0
E0
E0
E0
Some uses for Polarization
Polarized Sun Glasses
•“Glare” comes mostly from light scattered in the atmosphere and
reflected from water
•Mostly polarized
•Sun glasses use selective absorption to eliminate it
Optical Activity
•Some materials are capable of rotating the plane of polarization
•These materials are not mirror-symmetric
•Enantiomers, especially biological molecules
•Studying rotation of polarized light detects presence of these molecules
•Someday use these to detect life on other planets?
E0
Sugar
water
E0
CT -5 - When a ray of light is incident on two polarizers with
their polarization axes perpendicular, no light is transmitted. If a
third polarizer is inserted between these two with its polarization
axis at 45° to that of the other two, does any light get through to
point P?
A. yes
B. no
Ans A
Ex - Serway 38-43. Plane-polarized light is incident on a single
polarizing disk with the direction of E0 parallel to the direction
of the transmission axis. Through what angle should the disk be
rotated so that the intensity in the transmitted beam is reduced
by a factor of (a) 3 (b) 5, (c) 10?
Solve
on
Board