Download Optical interference component irradiances."

“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."
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.
Consider now linearly polarized plane waves arriving at point P (first slide).
The two waves can be described in the usual manner:
(
r r
r r
r
E1 (r , t ) = E01 cos k1 ⋅ r − ωt + ε 1
)
(
r r
r r
r
E2 (r , t ) = E02 cos k 2 ⋅ r − ωt + ε 2
and
A point P, the irradiance is given by I = εv E 2
since the medium doesn’t change.
(
)(
T
or simply
)
I = E2
)
T
r r
r r
r r
r r
2
2
Then E = E ⋅ E = E1 + E2 ⋅ E1 + E2 = E1 + E2 + 2 E1 ⋅ E2
2
⇒ I = I1 + I 2 + I12
where I1 = E
2
1 T
, I2 = E
2
2 T
r r
, I12 = 2 E1 ⋅ E2
The last term, I12, is referred to as the interference term, it can be evaluated as
follows:
(
) (
(
r r
r r
r r
r r
E1 ⋅ E2 = E01 ⋅ E02 cos k1 ⋅ r − ωt + ε 1 cos k 2 ⋅ r − ωt + ε 2
r r
r r
r r
= E01 ⋅ E02 cos k1 ⋅ r + ε 1 cos ωt + sin k1 ⋅ r + ε 1 sin ωt
r r
r r
× cos k 2 ⋅ r + ε 2 cos ωt + sin k 2 ⋅ r + ε 2 sin ωt
[ (
[ (
)
)
(
)
)
]
]
)
T
Consider now the effect of time averaging:
f (t )
T
1
=
T
t +T
∫ f (t ' )dt '
t
Since cos 2 ωt
= 1 / 2; sin 2 ωt = 1 / 2; 2 sin ωt cos ωt T = sin 2ωt T = 0
T
rT r
r r
r r
r r
r r
r r
E01 ⋅ E02
cos k1 ⋅ r + ε 1 cos k 2 ⋅ r + ε 2 + sin k1 ⋅ r + ε 1 sin k 2 ⋅ r + ε 2
⇒ E1 ⋅ E2 =
T
2
r r
r r
r r
E01 ⋅ E02
=
cos k1 ⋅ r + ε 1 − k 2 ⋅ r − ε 2
2
r r r r
r r
r r
Therefore I12 = 2 E1 ⋅ E2 = E01 ⋅ E02 cos δ
where δ = k1 ⋅ r − k 2 ⋅ r + ε 1 − ε 2
[ (
(
) (
)
(
) (
)
T
This is the Interference Term
Phase Difference
δ Contains the combined path length and phase angle difference.
r
r
Note that E01 ⊥ E02 ⇒ I12 = 0
and
I = I1 + I 2
Thus, the resultant wave at point P in this case can be linearly, circularly, or
elliptically polarized, but the total flux density (irradiance) won’t change.
)]
Consider the most common situation:
r
r
E01 || E02
and
⇒
E012
E022
⇒ I12 = E01 E02 cos δ = 2 I1 I 2 cos δ , Since I1 =
, I2 =
2
2
I = I1 + I 2 + 2 I1 I 2 cos δ (1)
Total Constructive
I max = I1 + I 2 + 2 I1 I 2 , δ = 0, ± 2π , ±4π , ... ⇒
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, ... ⇒ Total Destructive Interference,
which leads to
I min = I1 + I 2 − 2 I1 I 2 ,
r
r
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 Spherical E-M Waves
Note that the Eq. I = I1 + I 2 + 2 I1 I 2 cos δ (1) also holds for
spherical waves in which
r
r
r
r
E1 (r1 , t ) = E01 (r1 ) exp[i (kr1 − ωt + ε 1 )] and 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).
r
r
If E01 = E02 ; I1 = I 2 = I 0
1
[k (r1 − r2 ) + (ε1 − ε 2 )]
2
⇒ max in I occur when δ = 2πm, m = 0, ±1, ±2, ...
and
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).
Young’s two slit experiment in which
r
r
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.
ym
S1 B = r1 − r2 = a sin θ ≈ aθ , θ ≈
where S1 B is the
,
s
ym
mλ
s
r1 − r2 = mλ ≈ aθ ≈ a
; ym ≈ mλ
⇒ θm ≈
s
a
a
s
s
s
Δy = ( ym +1 − ym ) = (m + 1)λ − mλ = λ
a
a
a
⇒ Red fringes have a larger spacing and are broader on the
screen than blue fringes.
OPD
OPD = Optical
Path Length
Difference and
m = 0, 1, 2, 3…
for maxima
(bright fringes)
ya for two overlapping spherical
waves.
s
δ k (r1 − r2 ) π ya
I = 4 I 0 cos 2 (δ / 2 ), where
=
=
λ s
2
2
Detector
⎛ π ya ⎞
⇒ I = 4 I 0 cos 2 ⎜
⎟
Coherent Source
⎝λ s ⎠
Since δ = k (r1 − r2 ) and
r1 − r2 ≈
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 Δy α 1/a.
Notice how the width
and spacing of the
intensity distribution
depend on λ.
Other wave-front splitting interferometers: Fresnel’s Double Mirror
Cylindrical wavefront and the law of reflection gives:
SA = S1 A, SB = S 2B, SA + AP = r1 , SB + BP = r2
Again, Δy ≈ sλ / a
Difference in OPL
= r1 – r2 = mλ
s = distance between the plane containing S1 and S2
and the screen, just as in the Young’s Experiment.
s
Fresnel’s Double Prism ⇒
Again, Δy ≅
a
s
λ
a
a = 2d(n-1)α where α is the prism angle.
s is the distance between the source (virtual coherent
sources) and the screen, as shown in (a).
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 Δϕ = ± π ⇒ δ = k(r1-r2) ± π
Therefore, I = 4 I cos 2 δ = 4 I cos 2 ⎛⎜ πay ± π ⎞⎟ = 4 I sin 2 ⎛⎜ πay ⎞⎟; Δy ≈ s λ
0
0
0
λ
s
a
2
2
⎝ λs ⎠
⎝
⎠
Note the spatial shift in the fringe pattern.
Δy ≈
s
λ
a
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
)
nf
d
n1
OPL difference : Λ = n f AB + BC − n1 AD
AB = BC = d / cos θ t ⇒ Λ =
AD = AC sin θ i = AC
Therefore, Λ =
2n f d
cos θ t
nf
n1
2n f d
cos θ t
sin θ t =
Snell ' s Law : n1 sin θ i = n f sin θ t
− n1 AD
2n f d sin 2 θ t
where
n1 cos θ t
(1 − sin θ ) = 2n d cosθ
2
t
f
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 Δϕ = π for the two reflected beams.
Consider again the rope analogy:
Glass (n = 1.5)
Δϕ = 0
Air (n = 1)
Δϕ = π
Air (n = 1)
Δϕ = 0
Glass (n = 1.5)
Δϕ = 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.
Fig. 9.21 Circular
Haidinger fringes
centered on the lens
axis
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
⇒ Δx =
2α
λf
2α
and
with λ f =
λ0
nf
2 d m = ( m + 1 / 2) λ f
x
d
Where Δx 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
2
The geometry shows
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
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 Δd= -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
d
10 −2 m
=
= 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 θ 4
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 ,
2d cos θ p = (m0 − p )λ0
where m0 =
For θ p small , cos θ p ≅ 1 −
⇒ 2d
2
= pλ0
λ0
⎛ 2d
⎞
⇒ 2d cos θ p = ⎜⎜
− p ⎟⎟λ0
⎠
⎝ λ0
⇒ 2d (1 − cos θ p ) = pλ0
θ p2
2d
Note that θ m ≡ θ p and m = m0 − p
θ p2
2
⇒ θp =
pλ0
d
which is the inclination
angle of the pth fringe.
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 + ...
E0trt ′
1
( geometric series ) ⇒ E0 r = E0 r −
=0
⇒ 1 + r + r + ... =
2
2
1− r
1− r
2
4
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
⇒ 1 − r + r − ... =
( geometric series )
2
1+ r
⎡ 1− r2 ⎤
E0trt ′
2r
tt ′ ⎤
⎡
=
+
=
+
=
⇒ E0 r = E 0 r +
1
1
E
r
E0
E
r
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
(
⎛ E02 ⎞
⎜
⎟⎟
2 ⎜
⎝ 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
~
~
~
iω t
i (ωt −δ )
E1r = E0 re , E2 r = E0tr ′t ′e
, E3r = E0tr ′3t ′e i (ωt − 2δ )
~
~ ~
~
~
~
( 2 N −3)
i [ωt − ( N −1)δ ]
′
′
.... E Nr = E0tr
te
,
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 ⎢
2 − iδ ⎥
−
r
e ⎦
1
⎣
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
(
(
[2r / (1 − r )] sin (δ / 2) ;
=I
1 + [2r / (1 − r )] sin (δ / 2 )
2
2
Ir
)(
)(
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:
~
~
~
E1t = E0tt ′e iωt , E2t = E0tt ′r ′2 e i (ωt −δ ) , E3t = E0tt ′r ′4 ei (ωt − 2δ )
~
~ ~
~
~
~
.... E Nr = E0tt ′r ′2 ( N −1)e i [ωt −( N −1)δ ] ,
Et = E1t + E2t + E3t + ... E Nt
~ ~*
2
′
′
(
)
E
E
t
t
t
t
~
⎡
⎤
t t
Et = E0 eiωt ⎢
and
I
=
= Ii
t
2 − iδ ⎥
2
1 + r 4 − 2r 2 cos δ
⎣1 − r e ⎦
1
2
′
Since tt + r = 1 ⇒ I t = I i
2
2
1 + 2r / 1 − r
sin 2 (δ / 2)
1 − cos δ
From sin 2 (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 ≡⎜
2 ⎟
1
r
−
⎝
⎠
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.
r2
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 Δϕ = 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.
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