Chapter 8 Wave Optics (1) (May 15, 2007) 光的干涉 Wave optics (part 1) 1.The corpuscular (...

Chapter 8

Wave Optics (1)

(May 15, 2007)

光的干涉

Wave optics (part 1)1. The corpuscular (微粒 ) theory of light (Until t

he middle of 17th century , Newton (1642-1727) supported).

2. Ray optics can explain many of the properties of light, but there exist many other interesting and beautiful effects that cannot be explained by the geometric optics.

For example, why is the sky blue? What causes the beautiful colors in a soap bubble or an oil film? Why are the clouds and spoondrifts (浪花 ) at sea white?

3. Hygens-Fresnel principle: each section of a wavefront in the diffracting aperture (孔 ) is the source of a spherical wavelet. The amplitude of the light wave at any point beyond the aperture is the superposition of all these wavelets.

4. Interference effects of light were first observed by Thomas Yong in 1800 and the wave theory was formed by 1820s

8.1 Interference (干涉 ) of light

• Interference of wave motion, What is the phenomenon?

• Two coherent waves should be satisfied with the three conditions: what are they?

(1) the same frequency

(2) the same vibrational direction

(3) the same initial phase or fixed phase change.

光波

Light is also a part of electromagnetic waves. The light interference follows the rule of wave interference.

is permittivity (介电常数 ) and is permeability (磁导率 ).

光干涉的必要条件

A chart of electromagnetic spectrum

Visible light (very approximately):

400~450 nm Violet

450~500 nm Blue

500~550 nm Green

550~600 nm Yellow

600~650 nm Orange

650~700 nm Red

8.1.1 Optical length (optical path, distance，光程 )

It is known that the phase difference of two waves at their meeting point is expressed as

21

12 2xx

Light travels at a different speed in different medium.

What does that the speed of light in a medium depends on?

n

cv

What is the relation of wavelengths in medium and free space?

nn

cTvT 0

Where 0 is wavelength of light in free space. So the wavelength of light becomes shorter in medium.

Optical path is defined as the product of refractive index n and the geometrical distance d the light travels.

Optical Path dn

At the same period t, does light travel the same distance in two different medium?

In medium:

optical path = cttn

cnvtnLn

S

L n

n0

In free space:

Optical Path ctctnSn 00

Therefore, they have the same optical path.

The phase difference in medium n should be calculated as: (suppose 1 - 2 = 0)

000

22

/

2ndd

nAgain 0 is wavelength of light in free space, d is the geometrical length in the medium.

dxx

2

2 21

相干光

8.1.2 Young’s double-slit experiments

20~100cm 1~5m

Slits are 0.1~0.2 mm wide, separation of two slits < 1mm

1. Coherent conditions of light:

(1)The light wave has fixed wavelength. The incident beam should be monochromatic (单色的 ).

(2)The secondary wavelets that originate from the two small openings are in phase at their point of origin in the openings.

(3)The openings are small in comparison with the wavelength of the incident light.

(4)The distance between the two openings is not too big compared with the wavelength of the incident light.

2. Yong’s formulas for bright and dark fringes

In the above figure, S1P = BP, the light path difference is S2B = . S1S2 = d, Therefore,

L

xd

L

OPddd tansin

d

P

OS0

L

S1

S2B

xD

S2

S1

B

AC

D

• what is the constructive conditions for two waves?

• by phase difference

• by path difference if initial phase difference is zero

• what is the initial phase difference in double-slit interference experiment ?

•how about total phase change? What does it depend on?

Total phase change is therefore caused by the light path difference.

• Constructive interference

According to the interference theory of wave motion, whenever the path difference is an integer multiple of the wavelength, = m, the constructive interference or reinforcement interference should occurs as long as light is wave. Therefore,

md sin

mL

xd

For bright fringes:

(m = 0, 1, 2, …)

∴d

m sin

• Destructive interference

On the other hand, the opposite phenomenon occurs that the two light waves are cancelled each other. This condition is called destructive interference or cancellation and in this case, what is the optical path difference should be equal to?

Ld

mxm

)(sin 21 md (m = 0, 1, 2, …)

)( 21 m

L

xd m 2

1m

1sin m

d

So for dark fringes:

Lmd

xm

2

1

• The spacing of two bright or dark fringes:

Ld

Ld

mL

d

mxxx mm

)1(1

Yong’s experiments show that all the above formulas can describe the phenomena observed in his experiments very well, so the wave property of light is proved.

• The analysis of the results

1. The spacing between two dark or bright fringes is independent from m, so they are equally spaced.

2. As is small, so d cannot be too big, otherwise, they cannot be distinguished.

3. What will you get if you use the sun (white) light as a light source?

Ld

xxx mm

1

Example 1: In an interference pattern from double slits, the seventh-order bright fringe is 32.1mm from the zeroth-order bright fringe. The double slit is 5 meters away from the screen, and the two slits are 0.691mm apart. Calculate the wavelength of the light.

Solution: the data we know are

x7 = 3.21×10-2 m, d = 6.91×10-4 m, m = 7, L = 5 m, So we have:

nmmL

dxL

d

mx mm 634

57

1091.61021.3 42

Light interference gives us an important method for measuring the wavelength of light.

The contacting place of screen and mirror will always have dark fringe. This explains that the incident and reflected lights always have phase difference of .

•It occurs when the light wave initially traveling in an optically thinner medium ( 光疏介质 ) is reflected by an interface with an optically denser medium (光密介质 ).

• It is found that the reflecting light has a π phase change which is called abrupt phase change (相位突变 ).

• This phenomenon is called half-wavelength lost.

8.1.4 Interference in thin films

The colored bands in the reflection of light from a thin film of oil on water and the colors on the reflection of light from a soap bubble are common phenomenon. The interference of light rays reflected from opposite surfaces of a thin transparent film.

d

12

na

b

c

ObserverRay 1 and ray 2 produce the interference. The optical path difference of 1 and 2 depends on the thickness of the film.

Ray 1 has an abrupt phase change at point a where the light initially travels in an optically thinner medium and is reflected at the interface with an optically denser medium. The phase change of occurs at the upper surface of the film.

It is supposed that the direction of incident light is more or less perpendicular to the film surface. So the ray 2 has an extra optical path of approximate n*2*d and ray 1 lost half-wavelength because of reflection. Therefore, we have

2

22

22

212

ndnd

)2(22

, 21 dnnL

Therefore,

The difference of optical length is

22

nd

This explains that why the abrupt phase change has a special relation with the half-wavelength lost.

The condition of destructive interference is:

2

1

22 mnd

That is

The condition for constructive interference is

mnd 2

2 m = 1, 2, …

n

mdmnd

22

m = 1, 2, …

m = 1, 2, …)2

1(2 mnd

Example 1: A soap bubble 550nm thick and of refractive index 1.33 is illustrated at near normal incidence by white light. Calculate the wavelengths of the light for which destructive interference occurs.

Solution: what is the condition for destructive interference in such a case? *

)(10463.155033.122 3

nmmmm

nd

)(3664

)(488,3

)(732,2

)(1463,1

tultraviolenmm

greenbluenmm

rednmm

infrarednmm

mnd 2 m = 1, 2, …

In the visible region, the light from both ends of the spectrum is reflected with destructive interference. We can not see these wavelengths of visible light. The wavelengths we can see have to be calculated using the constructive condition of interference.

)(12

29262 ,2

212

1 nmmm

ndmnd

et)(ultraviol 325,5

)violet( 418,4

;yellow585,3

(IR) 975,2;2926,1

nmm

nmm

nmm

nmmnmm

）（

8.1.5 Equal thickness interference,

Generally speaking, the abrupt phase change occurs at one of the surface of the wedge. So it is easy to get the difference of optical length.

mnem 2 m = 0, 1, 2, …

The condition for destructive interference is simpler

22

neGlass plate

e

Zero-order dark fringe

Incident rayInterfering rays

Air wedge

nee

eL k 2

e ,sin 1k

nnL

2sin2

Example 2: two microscope slides each 7.5cm long are in contact along one pair of edges while the other edges are held apart by a piece of paper 0.012mm thick. Calculate the spacing of interference fringes under illumination by light of 632nm wavelength at near normal incidence.

Solution: let the air thickness em corresponding the mth-order dark fringe and em+1 to the (m+1)th-order dark fringe. As the refractive index of air is 1, we can write out:

2em = m, 2em+1 = (m+1) It is easy to find the spacing of two neighbor fringes by deleting m from above equations. We have

emB

22

)1(1

mmee mm

From the rules for similar triangles, we know

mmeeBC

ACx

ee

x

BC

AC

mm

97.1)( 11

0.012mm

Zero-order dark fringe

em+1

x

7.5cm

A

C

Newton’s rings

If the convex surface of a lens with large radius is placed in contact with a plane glass plate, a thin film of air is formed between the two surfaces. The thickness of this film is very small at the point of contact, gradually increasing as one proceeds outward. The loci (locus, 轨迹 ) of points of equal thickness are circles concentric with point of contact.

At such a case, the difference of optical length is

22

ne

Where e is the thickness of air film, /2 is from the half-wavelength lost for the two rays considered. The condition for bright fringes is

The condition for bright fringes is

12

2

1

22 kekne kk

The condition for dark fringes is

n

kekne k 22

)12(2

2

On the other hand, we could also calculate the radii of bright and dark rings.

e

R

r

R-e

O

C

2

222

2

)(

kk

kk

eRe

eRRr

As R>>e, e2 can be dropped. So we have:

The radius for kth bright ring is

... 2, 1, k 2

)12(

Rkrk

The radius for kth dark ring is

... 2, 1, k kRrk