Download “ waves, yielding a resultant irradiance that may deviate from the... component irradiances. r

Interference
“Optical interference corresponds to the interaction of two or more light
waves, yielding a resultant irradiance that may deviate from the sum of the
component irradiances.
(Interference of almost plan waves)
A distance r far from the two point sources, the spherical wave fronts arriving at
point P can be approximated as plane waves, as shown.
Interference of spherical waves
Consider now linearly polarized plane waves arriving at point P (first slide).
The two waves can be described in the usual manner:
(
E1 (r , t ) = E01 cos k1 ⋅ r − ωt + ε 1
)
A point P, the irradiance is given by
since the medium doesn’t change.
(
(
E2 (r , t ) = E02 cos k 2 ⋅ r − ωt + ε 2
and
)(
I = εv E 2
T
or simply
I ∝ E2
)
T
)
Then E 2 = E ⋅ E = E1 + E2 ⋅ E1 + E2 = E12 + E22 + 2 E1 ⋅ E2
I = I1 + I 2 + I12
where I1 = E12 , I 2 = E22 , I12 = 2 E1 ⋅ E2
T
T
The last term, I12, is referred to as the interference term, it can be evaluated
as follows:
(
) (
)
= E ⋅ E [cos(k ⋅ r + ε )cos ωt + sin (k ⋅ r + ε )sin ωt ]
× [cos(k ⋅ r + ε )cos ωt + sin (k ⋅ r + ε )sin ωt ]
E1 ⋅ E2 = E01 ⋅ E02 cos k1 ⋅ r − ωt + ε 1 cos k 2 ⋅ r − ωt + ε 2
01
02
1
2
1
2
1
2
1
2
T
Consider now the effect of time averaging:
Since cos 2 ωt
E1 ⋅ E2
=
T
=
T
= 1 / 2; sin 2 ωt
[ (
f (t )
T
1
=
T
t +T
f (t ')dt '
t
= 1 / 2; 2 sin ωt cos ωt
T
) (
)
T
= sin 2ωt
(
T
=0
) (
E01 ⋅ E02
cos k1 ⋅ r + ε 1 cos k 2 ⋅ r + ε 2 + sin k1 ⋅ r + ε 1 sin k 2 ⋅ r + ε 2
2
(
E01 ⋅ E02
cos k1 ⋅ r + ε 1 − k 2 ⋅ r − ε 2
2
Therefore I12 = 2 E1 ⋅ E2
T
)
= E01 ⋅ E02 cos δ
This is the Interference Term
where δ = k1 ⋅ r − k 2 ⋅ r + ε 1 − ε 2
Phase Difference
δ Contains the combined path length and phase angle difference.
Note that E01 ⊥ E02
I12 = 0
and
I = I1 + I 2
Thus, E-M waves with orthogonal state of polarization do not interfere.
)]
Consider the most common situation:
E01 || E02
and
I12
2
2
E01
E02
, I2 =
= E01 E02 cos δ = 2 I1 I 2 cos δ , Since I1 =
2
2
I = I1 + I 2 + 2 I1 I 2 cos δ
(1)
I max = I1 + I 2 + 2 I1 I 2 , δ = 0, ± 2π , ±4π , ...
Total Constructive
Interference
There is still constructive interference when 0 < cosδ < 1 and leads to
I1 + I2 < I < Imax .
For 0 > cos δ > -1 this leads to destructive interference.
When cosδ = −1, δ = ±π, ±3π, ±5π, ...
which leads to
Total Destructive Interference,
I min = I1 + I 2 − 2 I1 I 2 ,
If E01 = E02 , I1 = I 2 = I 0 and
I = 2 I 0 (1 + cos δ ) = 4 I 0 cos 2
δ
2
Interference of two E-M waves with parallel states of
polarization
I = I1 + I 2 + 2 I1 ⋅ I 2 cos δ
I1=I2=I0
Interference
Intensity
2 n
Total Constructive
(E1+E2)2 4I0
(2n+1/2) > > (2n-1/2)
Partial Constructive I>I1+I2
I>2I0
(2n+1/2)
No interference
I=I1+I2
I=2I0
Partial destructive
I<I1+I2
I<2I0
Total destructive
(E1-E2)2 0
(2n+3/2) > >(2n+1/2)
(2n+1)
Conditions for interference
•
•
•
•
Two beams will interfere under the following conditions:
Temporal coherence
Spatial coherence
Monochromatic beams
Parallel state of polarization
In case of partial coherence and natural white light partial interference will
show up under certain conditions.
Interference of Spherical E-M Waves
Note that the Eq. I = I1 + I 2 + 2 I1 I 2 cos δ
spherical waves in which
E1 (r1 , t ) = E01 (r1 ) exp[i (kr1 − ωt + ε 1 )] and
(1)
also holds for
E2 (r2 , t ) = E02 (r2 ) exp[i (kr2 − ωt + ε 2 )],
in which r1 and r2 are radii of the spherical waves overlapping at P.
The phase shift in this case is δ = k(r1 – r2) + (ε1-ε2).
If E01 = E02 ; I1 = I 2 = I 0
and
1
[k (r1 − r2 ) + (ε1 − ε 2 )]
2
max in I occur when δ = 2πm, m = 0, ±1, ±2, ...
I = 4 I 0 cos 2
min in I occur when δ = πm, m = ±1, ±3, ±5, ...
Consider ε1 = ε2 , then r1 – r2 = 2πm/k = mλ (max) for m = 0, ±1, ±2, …
and r1 – r2 = πm/k = mλ/2 (min) for m = 0, ±1, ±3, …
These conditions define hyperbolas in the 2D x-y plane or hyperboloids of
revolution in 3D (See Fig. 9.3 as previously shown).
Wavefront-splitting Interferometers
1. Young experiment
A spherical spatially coherent wave was created by letting the light to pass through a
single slit. Two coherent sources were created by letting the spherical beam to pass
through narrow slits at
Young’s two slit experiment in which
E01 = E02 ; I1 = I 2 = I 0
and the two sources are coherent.
Geometry of Young’s Slit Experiment
A determination of conditions for constructive interference and appearance
of bright fringes: OPD = r1 – r2 = mλ
θm
Assume that a << s and y << s, i.e., a small angle approx.
Ymis the distance of the mth white fringe from the axis
S1 B = r1 − r2 = a sin θ ≈ aθ , θ ≈
ym
,
s
where S1 B is the OPD
mλ
s
; ym ≈ mλ
a
a
s
s
s
∆y = ( ym +1 − ym ) = (m + 1)λ − mλ = λ
a
a
a
r1 − r2 = mλ ≈ aθ ≈ a
ym
s
θm ≈
Red fringes have a larger spacing and are broader on the
screen than blue fringes.
OPD = Optical
Path Length
Difference and
m = 0, 1, 2, 3…
for maxima
(bright fringes)
Since δ = k (r1 − r2 )
I = 4 I 0 cos 2 (δ / 2 ),
π
I = 4 I 0 cos
λ
2
ya for two overlapping spherical
and r1 − r2 ≈
waves.
s
δ k (r1 − r2 ) π ya
where
=
=
λ s
2
2
Detector
ya
Coherent Source
s
Laser
A path length difference of one wavelength corresponds to m = ± 1 and the
first order maximum, on both sides of the
central maximum (m = 0).
Ideal irradiance
(intensity) vs distance
curve. The fringe
separation ∆y α 1/a.
Notice how the width
and spacing of the
intensity distribution
depend on λ.
2, Fresnel’s Double Mirror
Other wave-front splitting interferometers: Cylindrical wavefront and the law of
reflection gives:
Difference in OPL
SA = S1 A, SB = S 2 B, SA + AP = r1 , SB + BP = r2
= r1 – r2 = mλ
s = distance between the plane containing S1 and S2
Again, ∆y ≈ sλ / a
and the screen, just as in the Young’s Experiment.
s
3. Fresnel’s Double Prism
a = 2d(n-1)α where α is the prism angle.
Again, ∆y ≅
a
s
λ
a
s is the distance between the source (virtual coherent
sources) and the screen, as shown in (a).
3. Lloyd Mirror
Flat dielectric or metal mirror. Cylindrical wavefront coming from the source at
S. Note that at glancing incidence, θi ≈ 90°, the amplitude reflection coef. = -1
which yields a phase shift of ∆ϕ = ± π
δ = k(r1-r2) ± π
s
δ
πay π
πay
Therefore, I = 4 I 0 cos 2 = 4 I 0 cos 2
±
; ∆y ≈ λ
= 4 I 0 sin 2
2
λs 2
λs
a
Note the spatial shift in the fringe pattern.
s
∆y ≈ λ
a
Amplitude splitting Interferometer
1. Dielectric Films: Double beam interference.
Dielectric Films – Double beam interference
Creation of fringe pattern from two reflected beams with
fields E1r and E2r; ignore higher order beams for now.
This is also referred to as an amplitude splitting device.
The beams with fields E1r and E2r are considered as
coming from two coherent virtual sources behind the film.
(
n1
)
OPL difference : Λ = n f AB + BC − n1 AD
AB = BC = d / cos θ t
AD = AC sin θ i = AC
Therefore, Λ =
2n f d
cos θ t
Λ=
nf
n1
2n f d
cos θ t
sin θ t =
2
t
2n f d sin 2 θ t
where
n1 cos θ t
f
d
E1r
n1
E2r
Snell 's Law : n1 sin θ i = n f sin θ t
− n1 AD
(1 − sin θ ) = 2n d cosθ
nf
t
AC = 2d tan θ t ( from Fig .)
Let n1 = n2 = n (film is in a single medium)
n < nf (e.g., soap film in Air)
n > nf (e.g. Air film between sheets of glass)
In either case, there will be a relative phase shift of π upon reflection.
For incident angles up to ~30 ° (see Fig. 4.44), regardless of [E0r]⊥ or [E0r]|| we
can expect that ∆ϕ = π for the two reflected beams.
Consider again the rope analogy:
Glass (n = 1.5)
∆ϕ = 0
Air (n = 1)
∆ϕ = π
Air (n = 1)
∆ϕ = 0
Glass (n = 1.5)
∆ϕ = 0
Thus, δ = k0 Λ ± π =
4πn f
λ0
d cos θ t ± π =
4π
λ0
d n 2f − n 2 sin 2 θ i ± π
Maxima occur when δ = 2mπ ,
m = 0, 1, 2,....
4πn f
4π
2
2
2
d n f − n sin θ i − π = 2mπ or
d cos θ t − π = 2mπ
λ0
λ0
d cos θ t = (2m + 1)
λ0
4n f
= (2m + 1)
λf
4
or
2d cos θ t = (m + 1 / 2)λ f
Note that this is for the case of maxima in reflected light, but also minima in the
transmitted light.
The case for minima in reflected light and maxima in transmitted light is as
follows:
For δ = (2m ± 1)π
2d cos θ t = mλ f
4π
λf
d cos θ t − π = (2m − 1)π
m = 1, 2, 3,....
Note that if n1 > nf > n2 or n1 < nf < n2, the π - phase shift would not be present.
For an extended source, light will reach the lens
from various directions, and the fringe pattern
will spread out over a larger area of the film.
If the lens used to focus
the rays has a small
aperture, fringes will
appear on a small
portion of the film.
Only the rays leaving
the point source that are
reflected directly into
the lens will be seen.
Haidinger fringes
Fringes of equal inclination.. t is the dominant parameter.
The central fringe has the highest m
Fringes of Equal Thickness
They are used to determine the surface feature of optical elements.
The optical thickness is the dominant parameter rather then i
Optical thickness nfd can also be made to vary instead of θi.
Consider a wedge-shaped film, as in the previous slide:
d
= tan α ≈ α
d = xα
x
For small θ i , (m + 1 / 2)λ0 = 2n f d m = 2n f x m α
xm =
(m + 1 / 2)λ f
λf
∆x =
2α
2α
and
with λ f =
d
λ0
nf
x
2 d m = ( m + 1 / 2) λ f
Where ∆x is the horizontal spacing between fringes and dm is the
film thickness at various maxima.
Note that this is an odd multiple of a quarter wavelength, λf/4 and
2× λf/4 = λf/2 ×2π / λf = π + π (reflection) → 2π
Newton’s rings
Consider the example of Newton’s rings:
A lens is placed on an optically flat surface, as shown in Fig. 9-23.
x 2 = R 2 − (R − d ) = 2 Rd − d 2
The geometry shows
2
R >> d
this becomes x 2 ≅ 2 Rd
mth-order interference max. occurs when
2n f d m = (m + 1 / 2 )λ0
(m + 1 / 2)λ f R
Bright ring:
xm =
Dark ring:
xm = mλ f R
λf =
Beam
splitter
λ0
nf
Glass
2. Michelson Interferometer – Mirror Interferometer
Mirror Interferometers are used to enable two beam interference
Michelson Interferometer (see next slide)
1) Silvered surface of beam-splitter (BS) is towards right side (back side).
2) Beam which is reflected by M2 passes through glass of BS three times.
3) Beam which is reflected by M1 passes through glass of BS once.
We need to insert “compensator plate” C in arm OM1.
Exact Duplicate of BS without the silver coating.
Therefore, the difference in OPL is actual path length difference.
4) An additional phase term is present in the OM2 arm, due to an internal
reflection in the BS, whereas the OM1 wave has an external reflection at the BS.
Therefore 2dcosθm = mλ0
Conditions for destructive interference.
m = 0, ±1, ±2,…, d is the optical path length difference between the mirrors, and
θm is the inclination angle for a given order m. Note that 2dcosθm is the OPD
between the two beams (rays).
Fig. 9.25 A conceptual
rearrangement of the
Michelson interferometer.
Consider a ring for a fixed order m and the diagram of the next slide:
As M2 moves towards M1’, d decreases
cos θm increases
θm decreases
Rings shrink towards center of screen in order to preserve the relation that
2dcosθm = mλ0. One order disappears when ∆d= -λ0/2.
Very precise changes in length are made by counting the number of fringes
that disappear.
Example: Consider a yellow light source with λ = 589 nm. When M2 is
moved 1 cm, the number of fringes which shrink through the center point is
10 −2 m
d
=
= 33,956 fringes
−9
λ0 / 2 589 ×10 m / 2
(
)
A central dark fringe for θm = 0 is represented by 2d = m0λ0
For d fixed (d = const.) successive dark fringes are given by the
following relations on the next slide:
Dark Fringes: front
view with eyes.
y
x
y
2d cos θ p = (m0 − p)λ0
θ1 θ2 θ3 θ4
S1 ’
S1’’
S1’’’
S1’’’’
S1’’’’’
z
p=5
4 3 2 1 0 (θ0= 0)
x
Dark fringes with angle θp from several
virtual sources. Side view to give partial 3D
visualization.
For d fixed (d = const.) successive
dark fringes are given by
2d cos θ1 = (m0 − 1)λ0
(see previous slide for
labeling of pth fringe)
2d cos θ 2 = (m0 − 2)λ0
2d cos θ 3 = (m0 − 3)λ0 ,
where m0 =
2d cos θ p = (m0 − p )λ0
2d cos θ p =
2d (1 − cos θ p ) = pλ0
2d
2
= pλ0
λ0
2d
λ0
− p λ0
Note that θ m ≡ θ p and m = m0 − p
For θ p small , cos θ p ≅ 1 −
θ p2
2d
θ p2
2
θp =
pλ0
d
which is the inclination
angle of the pth fringe.
3. Multiple Beam Interference.
Arising from multiple internal reflections in a film
Difference in OPL between adjacent rays is given by Λ = 2nf dcosθt .
Need to consider various products of amplitude reflection coef: r (1st surface)
and r’ (2nd surface), and amplitude transmission coef: t (1st surface) and t’
(2nd surface).
Need to find expressions for
E1r, E2r, E3r, which are
E
E
r = 0 r , t = 0t
E0 i
E0 i
E1r = E0 r
E2 r = E0tr ′t ′
E3r = E0tr ′3t ′
Similarly, the transmitted wave
fields are given by:
E1t = E0tt ′, E2t = E0tr ′2t ′, E3t = E0tr ′4t ′
Time-Reversal Invariance: Important principle in Optics
If a process occurs, the reverse can also occur with reflection and refraction, as in
the figure below. The time reversal of (a) below gives (b), but (b) in forward time
must actually give (c). Self consistency between (b) and (c) requires that
E0i tt ′ + E0i rr = E0i
tt ′ + r 2 = 1 and
and
E0i rt + E0i tr ′ = 0
r′ + r = 0
Called Stokes Relations
These are useful for relating amplitude transmission and reflection coefficients
between opposites sides of dielectric films.
Regarding the figure with multiple beams, we can now sum the amplitudes.
Assume case (i) in which Λ= mλ (i.e., difference in OPL between adjacent
beams is an integral number of λ).
Therefore, E0 r = E0 r + E0tr ′t ′ + E0tr ′3t ′ + E0tr ′5t ′ + ...
(
)
(
and with r ′ = −r ,
)
E0 r = E0 r − E0trt ′ + E0tr 3t ′ + E0tr 5t ′ + ... = E0 r − E0trt ′ 1 + r 2 + r 4 + ...
1
1 + r + r + ... =
( geometric series )
2
1− r
2
4
E0trt ′
=0
E 0 r = E0 r −
2
1− r
from tt ′ = 1 − r 2 ( Stokes )
For case (ii) Λ = (m + ½)λ and signs alternate due to an extra ½ λ in the OPL.
Case (i) in which Λ= mλ
Case (ii) Λ = (m + ½)λ
Case (ii) Λ = (m + ½)λ
Therefore, E0 r = E0 r − E0tr ′t ′ + E0tr ′3t ′ − E0tr ′5t ′ + ...
and with r ′ = −r ,
(
)
E0 r = E0 r + E0trt ′ − E0tr 3t ′ + E0tr 5t ′ − ... = E0 r + E0trt ′ 1 − r 2 + r 4 − ...
1
( geometric series )
1 − r + r − ... =
2
1+ r
E0trt ′
tt ′
1− r2
2r
=
+
=
=
+
E0
E
r
E0 r = E 0 r +
E
r
1
1
0
0
2
2
2
2
1+ r
1+ r
1+ r
1+ r
2
4
from tt ′ = 1 − r ( Stokes )
2
and
E02r
4r 2
=
Ir =
2
1+ r 2
(
)
2
E02
2
Consider now a more general approach involving an arbitrary phase δ = k0Λ in
which 0 ≤ δ ≤ 2π. A complex representation is necessary. The reflected fields at
point P can be described by
~
~
~
iωt
i (ωt −δ )
E1r = E0 re , E2 r = E0tr ′t ′e
, E3r = E0tr ′3t ′ei (ωt − 2δ )
~
~ ~
~
~
~
( 2 N −3)
i [ωt − ( N −1)δ ]
.... E Nr = E0tr ′
,
t ′e
Er = E1r + E2 r + E3r + ... E Nr
The addition of the complex fields
can be described and visualized
using a phasor diagram:
The addition of the complex fields
can be accomplished with a similar
geometric series, as before:
(
− iδ
~ ~
~
~
~
iωt r 1 − e
Er = E1r + E2 r + E3r + ... E Nr = E0 e
1 − r 2 e − iδ
(
)(
)
)
)
E02 2r 2 (1 − cos δ )
E02
1 ~ ~ * E02 r 2 1 − e −iδ 1 − e iδ
I r = Er Er =
=
, Ii =
2 − iδ
2 iδ
4
2
2
2 1− r e 1− r e
2 1 + r − 2r cos δ
2
(
Ir
)(
[
2r / (1 − r )] sin (δ / 2 )
=I
;
1 + [2r / (1 − r )] sin (δ / 2 )
2
2
i
2
2
2
2
(
)
Note that for r << 1, I r ≈ 4 I i r 2 sin 2 (δ / 2 )
Thus, we reduce to the case of two-beam interference when the reflectance is
very small and the leading terms of the first two beams become dominant.
The transmitted fields can also be handled with the same complex method:
~
~
~
2 i (ωt −δ )
iω t
′
′
′
, E3t = E0tt ′r ′4 ei (ωt − 2δ )
E1t = E0tt e , E2t = E0tt r e
~
~ ~
~
~
~
.... E Nr = E0tt ′r ′2 ( N −1)ei [ωt −( N −1)δ ] ,
Et = E1t + E2t + E3t + ... E Nt
~ ~*
2
′
′
(
)
E
E
t
t
t
t
~
t t
and
I
Et = E0 eiωt
=
= Ii
t
2 − iδ
1− r e
2
1 + r 4 − 2r 2 cos δ
1
2
Since tt ′ + r = 1
It = Ii
2
2
1 + 2r / 1 − r
sin 2 (δ / 2)
1 − cos δ
2
(
)
From sin δ / 2 =
2
Also note that
I i = I r + I t when there is no absorption or loss of
(
[ (
)
)]
energy.
Note that δ = 2πm gives (It)max = Ii and (Ir)min = 0 since sin2(δ/2) = 0;
Likewise, when δ = (2m + 1)π
(It)min and (Ir)max since sin2(δ/2) = 1
Since δ = k0 Λ = k0 (2n f d cosθ t ) =
2d cos θ t = (m + 1 / 2)λ f
4π
λ0
n f d cos θ t = (2m + 1)π
for (I t )min , (I r )max
This is the identical condition obtained for only two beams (two reflected waves).
It is therefore also the same condition for a maximum in the reflected intensity for
multiple beams.
Define the coefficient of finesse F as follows:
2r
F≡
1− r 2
A(θ ) =
2
Ir
F sin 2 (δ / 2)
=
I i 1 + F sin 2 (δ / 2 )
1
1 + F sin 2 (δ / 2 )
and
It
1
=
I i 1 + F sin 2 (δ / 2)
This is the Airy Function, which is shown in
the next slide for the transmittance.
As r → 1, It/Ii is very small, except
within sharp spikes at δ = 2mπ.
Notice that when is small, Ir/Ii
approaches sin2δ/2 as previous
mentioned. This reduces to the
two beam case: I = 4I0cos2δ/2
with δ =k0(r1-r2). The difference
is due to an added ∆ϕ = π due to
reflection in the case of dielectric
films which cause the cosine to
become a sine. This is similar to
the case for Lloyds mirror.
r2
Note also that multiple beam
interference results in a
redistribution of the energy
density in comparison to the twobeam case, resulting in sharp
peaks or dips for r large.
It
Ii
Ir
Ii
The Fabry-Perrot interferometer
F>324
=2 m
R>0.9
"
!
#, t, d, n
A
then
It
T2
=
if
T = 1− R − A
I i (1 − R) 2 + 4 R sin 2 (δ / 2 )
It
A 2
1
) ⋅ A(θ )
= (1 −
A(θ ) =
(1 − R )
Ii
1 + F sin 2 (δ / 2 )
Airy
=2 m
"
2
F=
π F
2π
=
2δ1/ 2
2
"
1/2
1/2
1/2=2/F
Finesse
Fabry-Perot Spectroscopy
Fabry-Perot
!
"
#
FSR
!Finesse
nm
!
#/$#
!
!