Chapter 8 - 汕头大学医学院

1、Chapter 8,Wave Optics (1) (May 11, 2005),A brief summary to the last lecture,The structure of the eye,2. Vision of the human eye,vision of the human eye,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 (

2、farsighted); (3) Astigmatic eye (散光眼),4. Optical instruments used in medicine (1) Magnifying glass (2) Compound microscope (3) Fibrescope,Wave optics (part 1) The corpuscular (微粒) theory of light (Until the middle of 17th century , Newton (1642-1727) supported). Ray optics can explain many of the pr

3、operties 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. The wave theory of light (Huygens (1629-95) Interference effects of light were first observed by Thomas Yon

4、g 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 el

5、ectromagnetic spectrum,Visible light (very approximately): 400450 nm Violet 450500 nm Blue 500550 nm Green 550600 nm Yellow 600650 nm Orange 650700 nmRed,8.1.1 Optical length (optical length, distance, path),In Chapter 4, we learned that the phase difference of two waves are expressed as,Light is al

6、so a part of electromagnetic waves. The light interference follows the rule of wave interference as you know in Yongs 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 th

7、e definition of the refractive index, we obtain,From the definition of wavelength, which is the period of the wave multiplying the speed of the wave, we have,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 l

8、ength of the light travel and it is defined as the product of refractive index and the geometrical distance the light travels.,Lets 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 ind

9、ex n. in free space: opitacl length = n0 S = n0 c t = ct,In medium: optical length =,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

10、can use the formula directly,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,8.1.2 Youngs double-slit experiments,20100cm,15m,Slits are 0.10.2 mm wide, separation of two slits 1mm,1. Cohere

11、nt conditions of light: The wave has fixed wavelength. The incident beam should be monochromatic (单色的). The secondary wavelets that originate from the two small openings are in phase at their point of origin in the openings. The openings are small in comparison with the wavelength of the incident li

12、ght. The distance between the two openings is not too big compared with the wavelength of the incident light.,2. Yongs formulas for bright and dark fringes,In the above figure, S1P = BP, the light path difference is S2B = . Therefore,what is the constructive conditions for two waves? by phase differ

13、ence 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, whe

14、never 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,For bright fringes:,Destructive interference On the other hand, the opposite phenomenon occurs that the two light wav

15、es 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,(m = 0, 1, 2, ),So for dark fringes:,The spacing of two bright or dark fringes:,Yongs experiments show that all the above formula

16、s can describe the phenomena observed in his experiments very well, so the wave property of light is proved.,The analysis of the results,The spacing between two dark or bright fringes is independent from m, so they are equally spaced. As is small, so d cannot be too big, otherwise, they cannot be di

17、stinguished. 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

18、apart. Calculate the wavelength of the light. Solution: the data we know are x7 = 3.2110-2 m, d = 6.9110-4 m, m = 7, L = 5 m, So we have:,Light interference gives us an important method for measuring the wavelength of light.,8.1.3 Lloyds mirror,Lloyds mirror is an optical instrument for producing in

19、terference 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 phase change which is called abrupt phase change. This phenom

20、enon 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 fro

21、m 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.,Ray 1 and ray 2 produce the interference. The light path length difference

22、 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 suppose

23、d 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,The difference of light path length is,This explains that why the abrupt

24、 phase change has a special relation with the half-wavelength lost.,The condition of destructive interference is:,That is,The condition for constructive interference is,m = 1, 2, ,Example 1: A soap bubble 550nm thick and of refractive index 1.33 is illustrated at near normal incidence by white light

25、. Calculate the wavelengths of the light for which destructive interference occurs. Solution: what is the condition for destructive interference in such a case? *,m = 1, 2, ,In the visible region, the light from both ends of the spectrum is reflected with destructive interference. We can not see the

26、se wavelengths of visible light. The wavelengths we can see have to be calculated using the constructive condition of interference.,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

27、path length.,m = 0, 1, 2, ,Glass plate,e,Zero-order dark fringe,Incident ray,Interfering rays,The condition for destructive interference is simpler,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

28、 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 f