Chapter 8

Wave Optics (1) (May 11, 2005)

A brief summary to the last lecture

1. The structure of the eye

Cornea (角膜)　 (1.376)

Aqueous humor

　 (水状液，1.336)

Iris (虹膜)

Crystalline lens Retina

Yellow spot

Vitreous humor

The cross-section of human eye

Ciliary muscles (睫状肌)

Pupil

The eye is nearly spherical in shape & 1inch in diameter.

Optic nerve O

2. Vision of the human eye

vision of the human eye1

from 0.1 ~ 1.5

is the viewing angle and in unit of minute.

3. Corrective eyeglasses for visual defects

(1) Myopia (nearsighted);

(2) Hyperopic eye (farsighted);

(3) Astigmatic eye (散光眼 )

4. Optical instruments used in medicine

(1) Magnifying glass

(2) Compound microscope

(3) Fibrescope

f

25

21

25

ff

sM

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, Experiments show that light bends around corners.

3. The wave theory of light (Huygens (1629-95))

4. Interference effects of light were first observed by Thomas Yong in 1800.

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 constant phase change.

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 length, distance, path)

In Chapter 4, we learned that the phase difference of two waves are expressed as

21

12 2xx

Light is also a part of electromagnetic waves. The light interference follows the rule of wave interference as you know in Yong’s double-slit experiment.

It is known that light travels at a different speed in different medium. The speed of light in a medium depends on the refractive index of the medium. From the definition of the refractive index, we obtain

n

cv

From the definition of wavelength, which is the period of the wave multiplying the speed of the wave, we have

nn

cTvT 0

Where 0 is wavelength of light in free space. So the wavelength of light becomes shorter in medium. Light path length is not the geometrical length of the light travel and it is defined as the product of refractive index and the geometrical distance the light travels.

Let’s have a look at the light path length at the same period t when it travels in two different medium. We choose medium one is free space and medium two with refractive index n.

in free space:

opitacl length = n0 S = n0 c t = ct

In medium:

optical length = cttn

cnvtnLn

So it is found that though the lights traveling in different medium have different geometrical path (S and L) at the same period, they have the same light path length. So when we calculate the phase in medium n, we can use the formula directly

nL0

2

Again 0 is wavelength of light in free space, L is the geometrical length in the medium. Also we can use the similar formula to calculate the phase difference

0

2

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 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 = . Therefore,

L

xd

L

OPddd tansin

d

P

OS0

L

S1

S2B

x

• what is the constructive conditions for two waves?

• by phase difference

• by path difference

• what is the initial phase change 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 changes.

• 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:

∴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 light path difference should be equal to? 陈文灯

Ld

mxm

)(sin 21 md (m = 0, 1, 2, …)

)( 21 m

L

xd 2

11sin 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

Ld

xxx mm

1

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? 陈善源

Example 1: In an interference pattern from two 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.

8.1.3 Lloyd’s mirror

Lloyd’s mirror is an optical instrument for producing interference fringes. A slit is illuminated by monochromatic light and placed close to a plane mirror. Interference occurs between direct light from the slit and light reflected from the mirror.

L

• It is found that the reflecting light has a π phase change which is called abrupt phase change.

• This phenomenon is called half-wavelength lost.

• 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 ).

8.1.4 Interference in thin films

It is easy for us to see 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. These phenomenon shares a common feature, 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 light path length 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 light path length of approximate n*2*d and ray 1 lost half-wavelength because of reflection. Therefore, we have

2

22

22

2

ndnd

The difference of light path 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, …n

md4

)12(

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

21

21

nmmm

nd

mnd

;325,5

)(418,4;585,3

)(975,2;2926,1

nmm

violetnmmnmm

orangenmmnmm

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 light path length.

mne2 m = 0, 1, 2, …

Glass platee

Zero-order dark fringe

Incident rayInterfering rays

The condition for destructive interference is simpler

22

ne

Air wedge

ne

eL k 2

ee ,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 e corresponding the mth-order dark fringe and e1 to the (m+1)th-order dark fringe. As the refractive index of air is 1, we can write out:

2e = m, 2e1 = (m+1)

e B

It is easy to find the spacing of two neighbor fringes by deleting m from above equations. We have

21

ee From similar triangles, we

know

mmeeBC

ACx

ee

x

BC

AC97.1)( 1

1

0.012mm

Zero-order dark fringe

e1

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 light-path-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 k

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

)(

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