1

Chapter 35

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.

Interference

35-

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

Law of Refraction

35-

Index of Refraction:c

nv

Fig. 35-3

1 2 1 1

1 2 2 2

vt

v v v

11

22

sin (for triangle hce)

sin (for triangle hcg)

hc

hc

1 1 1

2 2 2

sin

sin

v

v

1 21 2

and c c

n nv v

1 1 2

2 2 1

sin

sin

c n n

c n n

Law of Refraction: 1 1 2 2sin sinn n

4

Wavelength and Index of Refraction

35-

Fig. 35-4

nn n

v v

c c n

nn

v c n cf f

n 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

11 1

1 1

Number of wavelengths in : n

L L Lnn N

n

22 2

2 2

Number of wavelengths in : n

L L Lnn N

n

2 22 1 2 1 2 1Assuming :

Ln Ln Ln n N N n n

2 1 1/2 wavelength destructive interferenceN N

hittThe image of an erect candle, formed using a convex mirror (f < 0), is always:

A. virtual, inverted, and smaller than the candle

B. virtual, inverted, and larger than the candle

C. virtual, erect, and larger than the candle

D. virtual, erect, and smaller than the candle

E. real, erect, and smaller than the candle

5

hittA 5:0-ft woman wishes to see a full length image of herself in a plane mirror. The minimum length mirror required is:

A. 5 ft

B. 10 ft

C. 2.5 ft

D. 3.54 ft

E. variable: the farther away she stands the smaller the required mirror length 6

question

Two thin lenses (focal lengths f1 and f2) are in contact. Their equivalent focal length is:

A. f1 + f2

B. f1f2/(f1 + f2)

C. 1=f1 + 1=f2

D. f1 /f2

E. f1(f1 f2)=f2 7

8

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

9

Diffraction

35-

Fig. 35-7

For plane waves entering a single slit, the waves emerging from the slit start spreading out, diffracting.

10

Young’s Experiment

35-

Fig. 35-8

For waves entering a two slit, the emerging waves interfere and form an interference (diffraction) pattern.

11

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 DL of the rays reaching that point.

Path Length Difference: sinL d

1235-

Fig. 35-10

if sin integer bright fringeL d

Maxima-bright fringes:

sin for 0,1, 2,d m m

if sin odd number dark fringeL d

Minima-dark fringes: 12sin for 0,1,2,d m m

1 1.51 dark fringe at: sinm

d

Locating Fringes

1 22 bright fringe at: sinm

d

13

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(wt+f)

Coherent sources: Phase f 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: f jitters randomly in time, no stable interference occurs

14

Intensity in Double-Slit Interference

35-Fig. 35-12

1 0 2 0sin and sinE E t E E t E1

E2

2 10 24 cosI I 2

sind

1 1 12 2 2minima when: sin for 0,1,2, (minima)m d m m

12

2maxima when: for 0,1,2, 2 sin

sin for 0,1,2, (maxima)

dm m m

d m m

avg 02I I

15

Fig. 35-13

Proof of Eqs. 35-22 and 35-23

35-

0 0

10 0 2

sin sin ?

2 cos 2 cos

E t E t E t

E E E

2 2 2 10 24 cosE E

22 21 1

02 220 0

4cos 4 cosI E

I II E

phase path length

difference difference

2phase path length2

difference difference

2sind

Eq. 35-22

Eq. 35-23

16

hitt

In a Young's double-slit experiment the center of a bright fringe occurs wherever waves from the slits differ in the distance they travel by a multiple of:

A. a fourth of a wavelength

B. a half a wavelength

C. a wavelength

D. three-fourths of a wavelength

E. none of the above

17

In general, we may want to combine more than two waves. For example, there may be more than two slits.

Procedure:

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-

E1E2

E3E4

E

18

Interference from Thin Films

35-

Fig. 35-15

12 ?

0

19

Reflection Phase Shifts

35-

Fig. 35-16

n1 n2

n1 > n2

n1 n2

n1 < n2

Reflection Reflection Phase ShiftOff lower index 0Off higher index 0.5 wavelength

20

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 ( l / n), to calculate the phase.

2

odd number odd number2 wavelength = (in-phase waves)

2 2 nL

½ wavelength phase difference to difference in reflection of r1 and r2

2

0

22 integer wavelength = integer (out-of-phase waves)nL

22

n n

12

2

2 for 0,1,2, (maxima-- bright film in air)L m mn

2

2 for 0,1,2, (minima-- dark film in air)L m mn

21

Film Thickness Much Less Than l

35-

If L much less than l, for example L < 0.1l, 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

22

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.

23

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

24

Fig. 35-20

Michelson Interferometer

35-

2= (number of wavelengths

in same thickness of air)

a

LN

1 22 2 (interferometer)L d d

1

2 (slab of material of thickness

placed in front of ) mL L

L M

2 2= = (number of wavelengths

in slab of material)

mm

L LnN

2 2 2- = = n-1 (difference in wavelengths

for paths with and without

thin slab)

m a

Ln L LN N

For each change in path by 1l, 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 l and n.