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Chapter 35
Interference
The concept of optical interference is critical to understanding many natural
phenomena, ranging from color shifting in butterfly wings to intensity patterns
formed by small apertures. These phenomena cannot be explained using
simple geometrical optics, and are based on the wave nature of light.
In this chapter we explore the wave nature of light and examine several key
optical interference phenomena.
35- 1
Light as a Wave
Huygen’s Principle: All points on a wavefront serve as point sources of
spherical secondary wavelets. After time t, the new position of the wavefront will
be that of a surface tangent to these secondary wavelets.
Fig. 35-2
35- 2
Law of Refraction
λ1
λ2
λ1 v1
→
=
t= =
v1 v2
λ2 v2
λ1
sin θ1 =
(for triangle hce)
hc
sin θ 2 =
sin θ1
=
sin θ 2
c
n1 =
v1
sin θ1
=
sin θ 2
Fig. 35-3
λ2
hc
(for triangle hcg)
λ1 v1
=
λ2 v2
c
Index of Refraction: n =
v
c
and n2 =
v2
c n1 n2
=
c n2 n1
Law of Refraction:
n1 sin θ1 = n2 sin θ 2
35- 3
Wavelength and Index of Refraction
λn v
v
λ
= → λn = λ → λn =
c
n
λ c
v
cn c
fn =
=
= = f
λn λ n λ
The frequency of light in a medium is the same
as it is in vacuum
Since wavelengths in n1 and n2 are different,
the two beams may no longer be in phase
L
L
Ln1
Number of wavelengths in n1: N1 =
=
=
Fig. 35-4
λn1 λ n1
λ
L
L
Ln2
Number of wavelengths in n2 : N 2 =
=
=
λn 2 λ n2
λ
Ln2 Ln2 L
Assuming n2 > n1: N 2 − N1 =
−
= ( n2 − n1 )
λ
λ
λ
N 2 − N1 = 1/2 wavelength → destructive interference
35- 4
Rainbows and Optical Interference
Fig. 35-5
The geometrical explanation of rainbows given in Ch. 34 is incomplete.
Interference, constructive for some colors at certain angles, destructive for
other colors at the same angles is an important component of rainbows
35- 5
Diffraction
For plane waves entering a single slit, the waves emerging from the slit
start spreading out, diffracting.
Fig. 35-7
35- 6
Young’s Experiment
For waves entering a two slit, the emerging waves interfere and form an
interference (diffraction) pattern.
Fig. 35-8
35- 7
Locating Fringes
The phase difference between two waves can change if the waves travel paths of
different lengths.
What appears at each point on the screen is determined by the path length difference
∆L of the rays reaching that point.
Path Length Difference: ∆L
= d sin θ
Fig. 35-10
35- 8
Locating Fringes
if ∆L = d sin θ = ( integer )( λ ) → bright fringe
Maxima-bright fringes:
d sin θ = mλ for m = 0,1, 2,K
Fig. 35-10
if ∆L = d sin θ = ( odd number )( λ ) → dark fringe
Minima-dark fringes:
d sin θ = ( m + 12 ) λ for m = 0,1, 2,K
 2λ 
−1  1.5λ 
m
1
dark
fringe
at:
sin
θ
=
=
m = 2 bright fringe at: θ = sin −1 

35- 9

d
d




Coherence
Two sources to produce an interference that is stable over
time, if their light has a phase relationship that does not
change with time: E(t)=E0cos(ωt+φ)
Coherent sources: Phase φ must be well defined and
constant. When waves from coherent sources meet, stable
interference can occur. Sunlight is coherent over a short length
and time range. Since laser light is produced by cooperative
behavior of atoms, it is coherent of long length and time
ranges
Incoherent sources: φ jitters randomly in time, no stable
interference occurs
35-10
Intensity in Double-Slit Interference
E2
E1 = E0 sin ω t and E2 = E0 sin (ω t + φ )
2π d
I = 4 I 0 cos2 12 φ
φ=
sin θ
λ
maxima when: 12 φ = mπ for m = 0,1, 2,K → φ = 2mπ =
→ d sin θ = mλ for m = 0,1, 2,K
E1
2π d
λ
sin θ
(maxima)
minima when: 12 φ = ( m + 12 ) π → d sin θ = ( m + 12 ) λ for m = 0,1, 2,K
(minima)
I avg = 2 I 0
Fig. 35-12
35-11
Proof of Eqs. 35-22 and 35-23
Eq. 35-22
E ( t ) = E0 sin ω t + E0 sin (ω t + φ ) = ?
E = 2 ( E0 cos β ) = 2 E0 cos 12 φ
E 2 = 4 E02 cos2 12 φ
I E2
= 2 = 4 cos2 12 φ → I = 4 I 0 cos2 12 φ
I 0 E0
Eq. 35-23
β + β =φ
Fig. 35-13
 phase   path length 
 difference   difference 

=

λ
2π
 phase  2π  path length 
 difference  = λ  difference 




φ=
2π
λ
( d sin θ )
35-12
Combining More Than Two Waves
In general, we may want to combine more than two waves. For eaxample,
there may be more than two slits.
Prodedure:
1. Construct a series of phasors representing the waves to be combined.
Draw them end to end, maintaining proper phase relationships between
adjacent phasors.
2. Construct the sum of this array. The length of this vector sum gives the
amplitude of the resulting phasor. The angle between the vector sum
and the first phasor is the phase of the resultant with respect to the first.
The projection of this vector sum phasor on the vertical axis gives the
time variation of the resultant wave.
E4
E
E1
E3
E2
35-13
Interference from Thin Films
φ12 = ?
θ ≈0
Fig. 35-15
35-14
Reflection Phase Shifts
n1
n1
n1 > n2
n1 < n2
n2
n2
Reflection
Reflection Phase Shift
Off lower index
0
Off higher index
0.5 wavelength
Fig. 35-16
35-15
Equations for Thin-Film Interference
Three effects can contribute to the phase
difference between r1 and r2.
1. Differences in reflection conditions
λ
2. Difference in path length traveled.
0
2
3. Differences in the media in which the waves
travel. One must use the wavelength in each
medium (λ / n), to calculate the phase.
Fig. 35-17
½ wavelength phase difference to difference in reflection of r1 and r2
odd number
odd number
2L =
× wavelength =
× λn 2 (in-phase waves)
2
2
2 L = integer × wavelength = integer × λn 2 (out-of-phase waves)
λn 2 =
λ
n2
2L = (m +
2L = m
λ
n2
1
2
)
λ
n2
for m = 0,1, 2,K (maxima-- bright film in air)
for m = 0,1, 2,K (minima-- dark film in air)
35-16
Film Thickness Much Less Than λ
r1
r2
If L much less than l, for example L < 0.1λ, than phase
difference due to the path difference 2L can be
neglected.
Phase difference between r1 and r2 will always be ½
wavelength → destructive interference → film will appear
dark when viewed from illuminated side.
35-17
Color Shifting by Morpho Butterflies and Paper Currencies
For the same path difference, different wavelengths
(colors) of light will interfere differently. For example,
2L could be an integer number of wavelengths for red
light but a half integer wavelengths for blue.
Furthermore, the path difference 2L will change when
light strikes the surface at different angles, again
changing the interference condition for the different
wavelengths of light.
Fig. 35-19
35-18
Problem Solving Tactic 1: Thin-Film Equations
Equations 35-36 and 35-37 are for the special case of a higher index film
flanked by air on both sides. For multilayer systems, this is not always the
case and these equations are not appropriate.
What happens to these equations for the following system?
L
r2
r1
n1=1
n2=1.5
n3=1.7
35-19
Michelson Interferometer
∆L = 2d1 − 2d 2 (interferometer)
∆Lm = 2 L (slab of material of thickness
L placed in front of M 1 )
2 L 2 Ln
Nm =
=
(number of wavelengths
λm
λ
in slab of material)
Na =
Fig. 35-20
N m -N a =
2 Ln
λ
2L
λ
(number of wavelengths
in same thickness of air)
2L 2L
−
=
( n-1) (difference in wavelengths
λ
λ
For each change in path by 1λ, the interference
pattern shifts by one fringe at T. By counting the fringe
change, one determines Nm- Na and can then solve for
L in terms of λ and n.
for paths with and without
thin slab)
35-20