about interference pattern

1. 1
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.
Optical Interference
35-

2. 2
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.
Light as a Wave
35-
Fig. 35-2

3. 3
Law of Refraction
35-
Index of Refraction:
c
n
v
=
Fig. 35-3
1 2 1 1
1 2 2 2
v
t
v v v
λ λ λ
λ
= = → =
1
1
2
2
sin (for triangle hce)
sin (for triangle hcg)
hc
hc
λ
θ
λ
θ
=
=
1 1 1
2 2 2
sin
sin
v
v
θ λ
θ λ
= =
1 2
1 2
and
c c
n n
v v
= =
1 1 2
2 2 1
sin
sin
c n n
c n n
θ
θ
= =
Law of Refraction: 1 1 2 2sin sinn nθ θ=

4. 4
Wavelength and Index of Refraction
35-
Fig. 35-4
n
n n
v v
c c n
λ λ
λ λ λ
λ
= → = → =
n
n
v c n c
f f
nλ λ 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
1
1 1
1 1
Number of wavelengths in :
n
L L Ln
n N
nλ λ λ
= = =
2
2 2
2 2
Number of wavelengths in :
n
L L Ln
n N
nλ λ λ
= = =
( )2 2
2 1 2 1 2 1Assuming :
Ln Ln L
n n N N n n
λ λ λ
> − = − = −
2 1 1/2 wavelength destructive interferenceN N− = →

5. 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
Rainbows and Optical Interference
35-
Fig. 35-5

6. 6
Diffraction
35-
Fig. 35-7
For plane waves entering a single slit, the waves emerging from the slit
start spreading out, diffracting.

7. 7
Young’s Experiment
35-
Fig. 35-8
For waves entering a two slit, the emerging waves interfere and form an
interference pattern.

8. 8
The phase difference between two waves can change if the waves travel paths of
different lengths.
Locating Fringes
35-
Fig. 35-10
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: sinL d θ∆ =

9. 935-
Fig. 35-10
( ) ( )if sin integer bright fringeL d θ λ∆ = = →
Maxima-bright fringes:
sin for 0,1,2,d m mθ λ= = K
( ) ( )if sin odd number dark fringeL d θ λ∆ = = →
Minima-dark fringes: ( )1
2sin for 0,1,2,d m mθ λ= + = K
1 1.5
1 dark fringe at: sinm
d
λ
θ −  
= =  ÷
 
Locating Fringes
1 2
2 bright fringe at: sinm
d
λ
θ −  
= =  ÷
 

10. 10
Coherence
35-
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

11. 11
Intensity in Double-Slit Interference
35-
Fig. 35-12
( )1 0 2 0sin and sinE E t E E tω ω φ= = +
2 1
0 24 cosI I φ=
2
sin
dπ
φ θ
λ
=
( ) ( )1 1 1
2 2 2minima when: sin for 0,1,2, (minima)m d m mφ π θ λ= + → = + = K
1
2
2
maxima when: for 0,1,2, 2 sin
sin for 0,1,2, (maxima)
d
m m m
d m m
π
φ π φ π θ
λ
θ λ
= = → = =
→ = =
K
K
avg 02I I=

12. 12
Fig. 35-13
Proof of Eqs. 35-22 and 35-23
35-
( ) ( )
( )
0 0
1
0 0 2
sin sin ?
2 cos 2 cos
E t E t E t
E E E
ω ω φ
β φ
= + + =
= =
β β φ+ =
2 2 2 1
0 24 cosE E φ=
2
2 21 1
02 22
0 0
4cos 4 cos
I E
I I
I E
φ φ= = → =
( )
phase path length
difference difference
2
phase path length2
difference difference
2
sind
π λ
π
λ
π
φ θ
λ
   
 ÷  ÷
   =
   
= ÷  ÷
   
=
Eq. 35-22
Eq. 35-23

13. 13
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.
Combining More Than Two Waves
35-
E1
E2
E3
E4
E

14. 14
Interference from Thin Films
35-
Fig. 35-15
12 ?φ =
0θ ≈

15. 15
Reflection Phase Shifts
35-
Fig. 35-16
n1 n2
n1 > n2
n1 n2
n1 < n2
Reflection Reflection Phase Shift
Off lower index 0
Off higher index 0.5 wavelength

16. 16
Equations for Thin-Film Interference
35-
Fig. 35-17
Three effects can contribute to the phase
difference between r1 and r2.
1. Differences in reflection conditions
2. Difference in path length traveled.
3. Differences in the media in which the waves
travel. One must use the wavelength in each
medium (λ / n), to calculate the phase.
2
odd number odd number
2 wavelength = (in-phase waves)
2 2
nL λ= × ×
½ wavelength phase difference to difference in reflection of r1 and r2
2
λ 0
22 integer wavelength = integer (out-of-phase waves)nL λ= × ×
2
2
n
n
λ
λ = ( )1
2
2
2 for 0,1,2, (maxima-- bright film in air)L m m
n
λ
= + = K
2
2 for 0,1,2, (minima-- dark film in air)L m m
n
λ
= = K

17. 17
Film Thickness Much Less Than λ
35-
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.
r2
r1

18. 18
Color Shifting by Morpho Butterflies and Paper Currencies
35-
Fig. 35-19
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.

19. 19
Problem Solving Tactic 1: Thin-Film Equations
35-
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?
n1=1 n2=1.5 n3=1.7
r1
r2
L