A phenomenon of far-reaching importance occurs when two waves travel along the same path and combine their strengths. It's not hard to see, particularly if one visualizes water waves, that two waves may either reinforce each other or cancel, depending on their relative positions.

Waves reinforcing (constructive interference)	Waves canceling (destructive interference)

Notice that two waves reinforcing each other may be made to cancel simply by shifting one of the waves by one-half £.

Such shifts are accomplished most commonly by reflecting waves at the interface between two media. There are only two types of reflection to be considered: One where the wave speeds up as it enters the second medium, the other where the wave slows down in the second medium. That is, the index of refraction is relatively less or greater in the second medium.

A mechanical analog is a heavy string attached to a light string.

	Initial:	Final:
Heavy string into light string
Light string into heavy string

In a like manner, light traveling from glass (high optical density, high n) to air (low density, relatively low n) has no phase reversal upon reflection. Light entering glass from air does undergo a phase reversal. Visualizing all this gets a little rough, so I will designate a Reversed and a Non-reversed phase as R and NR.

Only a small percentage of the light is reflected, but that is the part that most concerns us. The bulk of the energy goes right through the interface without any change in phase.

Look now at what happens when light enters a thin glass plate. Tiny amounts will be reflected or transmitted many times between the two sides of the glass.

Because so little energy survives the first one or two reflections, we will look at only the first reflections. This will greatly simplify the problem.

Look at what's happened. The two reflected rays (R and NR are shifted relative to one another. By adjusting the thickness of the glass plate, the amount of shift can be set to any desired amount. Depending on the thickness of the glass, the rays will reinforce each other and be seen, or interfere with each other and not be seen.

For a given thickness, certain wavelengths (colors) of light will reinforce while others will cancel. Thus a thin glass plate or a thin soap bubble will reinforce only a particular color for a particular thickness, but over a large surface will reflect a brilliant rainbow from incoming white light.

A note about light: As we saw in the last chapter, light is an electric (and magnetic) vibration in space. The frequency of that vibration determines whether the wave is a radio wave, light, an x-ray, or other electromagnetic wave. The wavelengths of visible light extend from about 400 nm to 700 nm, from violet to red. Since light is slowed upon entry into glass, its wavelength is reduced from £ to £/n. In the problems that follow assume the £ given or asked for is that within the thin layer.

The nanometer (nm) is a unit of length commonly used for light. 1 nm = 10^-9 m.

23. What thickness of glass will reflect two rays which will reinforce each other if the incoming light has a wavelength of 400 nm in glass?

One reflected wave is reversed, the other not reversed. To bring them into step, one wave must be shifted (1/2)£. This means the glass must be (1/4)£ thick--(1/4)£ down + (1/4)£ back makes a (1/2)£ delay. Hence: Thickness = £/4 Thickness = 400 nm/4 Thickness = 100 nm

24. What thickness of glass will NOT reflect light of 400 nm wavelength?

To keep the reversed and not reversed waves out of step, the ray is shifted a full wavelength. Thus the glass must be £/2 thick. T = £/2 T = 400 nm/2 T = 200 nm

In these problems we are assuming that the wavelengths refer to the wavelengths in glass, water, or air...whichever the thin layer consists of. For example, a 600 nm wavelength in air becomes a £ = 600/1.5 = 400 nm in glass of n = 1.5.

25. Two glass plates are placed together so as to leave a narrow air gap between them. What thickness of air gap will allow the reflection of 500 nm light?

To be in step, the two waves will have to be £/2 apart. Thus the air gap must have a thickness £/4. T = £/4 T = 500 nm/4 T = 125 nm

26. Oil has an optical density between air and water. What thickness of oil layer on water will result in no reflection of 600 nm light?

To get the two reversed rays out of step, one must be delayed £/2 (or 3£/2, 5£/2, etc..). Thus the oil must be £/9 thick. T = £/4 T = 600/4 T = 150 nm

I hope you've noticed an apparent inconsistency. If light strikes the film, yet none is reflected, where is the energy going? It turns out that when no light is reflected, 100% of the light is transmitted.

27. What thickness of oil on water would allow reflection of 6000 A light?

To get the reversed waves in step, one must be delayed £ (or 2£, 3£, ... n£). We will take the thinnest layer, so: T = £/2 T = 600/2 T = 300 nm

28. In a 400 nm thick oil layer, what two wavelengths will be reflected? What two will be transmitted?

To reflect, £ and 2£ (or n£) shifts of one ray will bring them into step. T = £1/2 T = £2 T = 400 nm £1 = 800 nm £2 = 400 nm To not refect, £/2 and 3£/2 shifts are needed. T = £1/4 T = 3£2/4 £1 = 1600 nm £2 = 533 nm

The term interference is used to describe the interaction of waves. Destructive interference is the cancellation on wave energy. Constructive interference is the reinforcement of waves. What we have been looking at interference from very thin layers of material is called thin film interference.

You have had contact with it in your own experience. Soap bubbles are colored because certain wavelengths of light will be reflected when the soap film reaches a particular thickness. It isn't the soapy water itself that is colored, merely the light reflected from it.

The same thing is true with oil films on water. After a rain you can see bull's-eye shaped blobs of rainbow colors on street pavement. These are drops of oil or gasoline that have gotten very thin. When light reflects off these layers, the color whose wavelength is just twice the thickness of the oil film will be reflected and seen. (The oil itself is not brilliantly colored.) If you investigate these blobs, notice that they always end up with white on the outside ring. Why is this true?

Iridescent colors in general are usually the result of this phenomenon. Peacock feathers get their brilliance from thin-film interference. Opals, beetle-backs, and some butterfly wings owe their colors to this phenomenon.

Another application comes with the coating of lenses. Like an oil film on water the coatings on lenses have intermediate optical density. No reflection of light will occur when the layer is 1/4 of the incoming wavelength. In optical equipment such as cameras and binoculars, reflections from the lens surfaces seriously harm the images formed. The lenses are coated to prevent reflections. Since preventing reflection improves transmission, a further benefit is gleaned from the procedure. The violet appearance of coated lenses comes from short-wavelength violet light which is reflected by the same layer designed to prevent reflection of yellow light the middle of the visible spectrum.

29. Light of wavelength l is to be reflected off a soap bubble. Find a general formula that will give all the thicknesses that will reflect the given wavelength.

Reflection will occur at shifts of £/2, 3£/2, (2n - 1)£/2 T = (1/2) shift T = (2n -1)£/4

30. A radar station is to transmit waves through a 58 cm thick plaster wall. What are the three longest waves that would be best transmitted?

Reflection will not occur when one ray is shifted n£. Thus: T = (1/2)n£ £ = 2T/n £ = 116 cm/n The three longest waves are: £ = 116/1, 116/2, 116/3 £ = 116 cm, 58 cm, 38.7 cm

We have considered waves traveling in only one dimension. Frequently in nature, waves travel in two or three dimensions and the resulting interference patterns become increasingly complex.

Consider water waves approaching a beach. If a breakwater is built with one hole in it, the waves will enter and spread out like like those in the image to the left.

If a second hole is made in the breakwater, waves will enter through both, and their amplitudes will combine.

Place yourself on the beach. If you stand at an equal distance from each opening, crests and troughs from each will arrive at the same time and reinforce each other. In other words, you will see waves at that point. (point A)

Walk a ways down the beach (point B). There will be some place that you will find yourself somewhat farther from one opening than the other. If this difference is just one-half wavelength, crests will meet troughs and there will be destructive interference. In other words, you will not see waves at that point. As you get farther and farther down the beach, the path difference through which waves must travel gets greater and greater. When there is a full wave difference, the waves reinforce. When there is a 1.5 waves difference, we again get destructive interference.

Don't get the idea that there is a special wave headed for point B or point A, there are waves headed in all directions at all times. Every point on the beach is receiving waves from both openings at all times. What we have done is simply recognize that at different points on the beach the waves from the two opening will have traveled different distances, and therefore may be out of step with each other.

31. Waves of 3 m wavelength pass through a breakwater. Donna Parrell stands on the beach, 35 m from one opening and 30.5 m from the other. Will she see waves at her location?

The difference in paths is: 35.0 - 30.5 = 4.5 m. This is 1.5£, so the waves arrive out of phase and will cancel. Donna will see no waves.

Because of the geometric messiness of the situations illustrated, we will shift our view to an easier set-up: we will move our beach (or whatever) very far away from the openings. We will move so far away that waves traveling from both sources will travel in essentially parallel paths to reach a point on the beach. (Obviously they won't be exactly parallel, but exactness is an illusion anyway. )

Now look at the picture:

note A: This is the difference between the lengths of the two paths note B: The two waves travel the same distance from here to the observer Study and ask questions about this diagram until it makes sense. Everything from here on out depends on the logic of this picture.

Now we'll put some values into the diagram:

Important equations: sin Ø = Çp/d Çp = dsinØ In the diagram, when Çp = 0, the waves are traveling straight ahead, and there is reinforcement (constructive interference). When Çp = 1/2 the waves arrive out of step and there is destructive interference. When Çp = 3.5, we again have destructive interference.

32. 3 m waves approach openings 11 m apart. At what angles will destructive interference occur?

For destructive interference: Çp = £/2, 3£/2, 5£/2, ... Çp = 3/2, 9/2, 15/2, ... sinØ = ÇP/d sinØ = (3/2)/11, (9/2)/11, (15/2)/11, ... Ø = 9.8°, 29°, 43°, ...

33. Light waves of 700 nm approach two slits separated by 2,000 nm. At what angles will the light rays be in phase? That is, where will constructive interference occur?

For reinforcing waves: Çp = 0, £, 2£, 3£, ... Çp = 0, 700 nm, 1400 nm, ... sinØ = 0, (700)/(2000), (1400)/(2000), ... Ø = 0, 20.5°, 44.4°, ...

34. Two speakers are 3.0 m apart. They produce sound of 0.8 m wavelength. At what angle will the third minimum occur? (That is, where will the third location of destructive interference be?)

Çp = 5£/2 Çp = [(5)(.8)]/(2) Çp = 2 sinØ = Çp/d sinØ = 2/3 Ø = 42°

Take note: the angle we have worked with is also the angle of the observer from the central maximum!

35. Sound of 580 hz frequency is emitted by two speakers separated by 2.8 m. How far from the central maximum will we find the first maximum?

Çp = £ sinØ = Çp/d sinØ = £/2.8 £ = v/f £ = 330/580 sinØ = (330/580)/2.8 Ø = 11.7°

36. Microwaves of unknown frequency are passed through double slits separated by 35 cm. The first maximum is found 15.8° from the center line. What is the wavelength of the microwaves?

Çp = £ sin(15.8°) = £/35 £ = 35sin15.8 £ = 9.5 cm

37. Light of 500 nm wavelength passes through double slits. The 3rd minimum is found 7.2° from the center line. What is the separation of the slits?

Çp3 = 5£/2 sin(7.2°) = (5£/2)/d sin(7.2°) = [(5 * 5000)/2]/d d = 9973 nm d = 9.97 * 10-6 m

38. Two speakers are separated by 1.3 m. The 2nd minimum occurs at 38° from the center line. What is the frequency of the sound?

Çp2 = 3£/2 sin(38°) = (3£/2)/1.3 £ = .53 m f = v/£ f = 330/.53 f = 618 hz