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Eðlisfræði 2, vor 2007

35. Interference

Assignment is due at 2:00am on Wednesday, January 17, 2007

Credit for problems submitted late will decrease to 0% after the deadline has passed.The wrong answer penalty is 2% per part. Multiple choice questions are penalized as described in the online help.The unopened hint bonus is 2% per part.You are allowed 4 attempts per answer.

A really simple introductory look at interference

A Simple Introduction to InterferenceLearning Goal: To understand the basic principles underlying interference.One of the most important properties of waves is the principle of superposition. The principle of superposition for waves states that when two waves occupy the same point, their effect on the medium adds algebraically. So, if two waves would individually have the effect "+1" on a specific point in the medium, then when they are both at that point the effect on the medium is "+2." If a third wave with effect "-2" happens also to be at that point, then the total effect on the medium is zero. This idea of waves adding their effects, or canceling each other's effects, is the source of interference.

First, consider two wave pulses on a string, approaching each other. Assume that each moves with speed meter per second. The figure shows the string at time . The effect of each wave pulse on the string (which is the medium for these wave pulses) is to displace it up or down. The pulses have the same shape, except for their orientation. Assume that each pulse displaces the string a maximum of meters, and that the scale on the x axis is in meters.

Part AAt time , what will be the displacement at point ?

Express your answer in meters, to two significant figures.

ANSWER: = 0

Part BChoose the picture that most closely represents what the rope will actually look like at time .

ANSWER: A B C D

The same process of superposition is at work when we talk about continuous waves instead of wave pulses. Consider a sinusoidal wave as in the figure.

[


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Part CHow far to the left would the original sinusoidal wave have to be shifted to give a wave that would completely cancel the original? The variable in the picture denotes the wavelength of the wave.

Express your answer in terms of .

ANSWER: =

Part DIn talking about interference, particularly with light, you will most likely speak in terms of phase differences, as well as wavelength differences. In the mathematical description of a sine wave, the phase corresponds to the argument of the sine function. For example, in the function , the value of at a particular point is the phase of the wave at that point. Recall that in radians a full cycle (or a full circle) corresponds to radians. How many radians would the shift of half a wavelength from the previous part correspond to?Express your answer in terms of .

ANSWER: phase difference = radians

Part EThe phase difference of radians that you found in the previous part provides a criterion for destructive interference. What phase difference corresponds to completely constructive interference (i.e., the original wave and the shifted wave coincide at all points)?Express your answer as a number in the interval .

ANSWER: phase difference = 0 radians

Part F

Since sinusoidal waves are cyclical, a particular phase difference between two waves is identical to that phase difference plus a cycle. For example, if two waves have a phase difference of , the interference effects would be

the same as if the two waves had a phase difference of . The complete criterion for constructive interference

between two waves is therefore written as follows:

Write the full criterion for destructive interference between two waves.

Express your answer in terms of and .

ANSWER: phase difference =

The phase for a plane wave is a somewhat complicated expression that depends on both position and time. For most interference problems, you will work at a specific time and with coherent light sources, so that only geometric considerations are relevant. Consider two light rays propagating from point A to point B in the figure,


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which are apart. One ray follows a straight path, and the other travels at a angle to that path and then reflects

off a plane surface to point B. Both rays have wavelength .

Part GFind the phase difference between these two rays at point B.

Part G.1 Find the difference in distanceFind the difference in length between the direct path and the reflected path. You can use the fact that triangle ABC is an equilateral triangle.Express your answer in terms of .

ANSWER: path length difference =

Now that you have the difference in path length, convert that to radians. Recall that every cycle of radians is equivalent to one wavelength.



Part HSuppose that the reflected ray receives an extra half-cycle phase shift when it reflects. What is the new phase shift at point B?

Hint H.1 How many radians in a half cycle?Hint not displayed



Whenever light reflects from a transparent interface, moving from lower index of refraction to higher index of refraction, it gets an extra half cycle phase difference. Being able to accurately find the phase differences between waves at various points will be useful in both interference and diffraction problems.

Introduction, some problems, and a couple of challenging problems on two slit interference

Understanding Two-Source InterferenceLearning Goal: To understand the assumptions made by the standard two-source interference equations and to be able to use them in a standard problem.

For solving two-source interference problems, there exists a standard set of equations that give the conditions for constructive and destructive interference. These equations are usually derived in the context of Young's double slit experiment, though they may actually be applied to a large number of other situations. The underlying assumptions upon which these equations are based are that two sources of coherent, nearly monochromatic light are


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available, and that their interference pattern is observed at a distance very large in comparison to the separation of the sources. Monochromatic means that the wavelengths of the waves, which determine color for visible light, are nearly identical. Coherent means that the waves are in phase when they leave the two sources.

In Young's experiment, these two sources corresponded to the two slits (hence such phenomena are often called two-slit interference). Under these assumptions, the conditions for constructive and destructive interference are as follows:

for constructive interference

,

and for destructive interference

,

where is the separation between the two sources, is the wavelength of the light, is an arbitrary integer, and is the angle between a line perpendicular to the line segment connecting the sources and the line from the midpoint of that segment to the point where the interference is being observed. These equations are often spoken of in terms of visible light, but they are, in fact, valid for any sort of waves, as long as the two sources fit the other criteria given.

Part AWhich of the following scenerios fits all of the criteria for the two-source interference equations to be valid?

ANSWER: Answer not displayed

Part BPart not displayed

Part CPart not displayed

Part DPart not displayed

Part EPart not displayed

Part FPart not displayed

FM Radio Interference

You are listening to the FM radio in your car. As you come to a stop at a traffic light, you notice that the radio signal is fuzzy. By pulling up a short distance, you can make the reception clear again. In this problem, we work through a simple model of what is happening.

Our model is that the radio waves are taking two paths to your radio antenna:


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the direct route from the transmitteran indirect route via reflection off a building

Because the two paths have different lengths, they can constructively or destructively interfere. Assume that the transmitter is very far away, and that the building is at a 45-degree angle from the path to the transmitter.

Point A in the figure is where you originally stopped, and point B is where the station is completely clear again. Finally, assume that the signal is at its worst at point A, and at its clearest at point B.

Part AWhat is the distance between points A and B?

Part A.1 What is the path-length difference at point A?Part not displayed

Part A.2 What is the path-length difference at point B?Part not displayed

Part A.3 What is the path length of reflected waves?Part not displayed

Express your answer in wavelengths, as a fraction.

ANSWER: = Answer not displayed wavelengths

Part BYour FM station has a frequency of megahertz. The speed of light is about meters per second. What is the distance between points A and B?

Express your answer in meters, to two significant figures.

ANSWER: = Answer not displayed

Double Slit 1

Two lasers are shining on a double slit, with slit separation . Laser one has a wavelength of , while laser two has a wavelength of . The lasers produce separate interference patterns on a screen a large distance away from the slits.

Part AWhich laser has its first maximum closer to the central maximum?

Hint A.1 Path differenceHint not displayed


Part B


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What is the distance between the first maxima (on the same side of the central maximum) of the two patterns?

Part B.1 Find the location of the first maximum for laser onePart not displayed

Part B.2 Find the location of the first maximum for laser twoPart not displayed



Part CWhat is the distance between the second maximum of laser one and the third minimum of laser two, on the same side of the central maximum?

Part C.1 Find the location of the second maximumPart not displayed

Part C.2 Find the value of the third minimumPart not displayed

Part C.3 Find the location of the third minimumPart not displayed



Double Slit 2

A laser with wavelength is shining light on a double slit with slit separation . This results in an interference pattern on a screen a distance away from the slits. We wish to shine a second laser, with a different wavelength, through the same slits.

Part AWhat is the wavelength of the second laser that would place its second maximum at the same location as the fourth minimum of the first laser?

Hint A.1 Second maximum and fourth minimumHint not displayed

Hint A.2 Locating interference maxima and minimaHint not displayed

Express your answer in terms of


Double Slit with Reflections

A radar tower sends out a signal of wavelength . It is meters tall, and it stands on the edge of the ocean. A weather balloon is released from a boat that is a distance out to sea. The balloon floats up to an altitude . In this


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problem, assume that the boat and balloon are so far away from the radar tower that the small angle approximation holds.

Part ADue to interference with reflections off the water, certain wavelengths will be weak when they reach the balloon. What is the maximum wavelength that will interfere destructively?

Hint A.1 Small-angle approximationHint not displayed

Hint A.2 Interference with reflectionsHint not displayed

Hint A.3 Double slit approximationHint not displayed

Express your answer in terms of , , and .


Part BWhat is the maximum wavelength that will interfere constructively?



Two-Slit Interference

As Richard Feynman stated in his book on quantum mechanics,

Interference contains the heart and soul of quantum mechanics.

In fact, interference is a phenomenon of classical waves, easily perceived with sound or light waves. (It contains the soul of quantum mechanics only after you swallow the preposterous notion that particles in motion are described by a wave equation rather than the laws of Newtonian mechanics.)

In this problem, you will look at a classic wave interference problem involving electromagnetic waves. Young's double-slit experiment provided an irrefutable demonstration of the wave nature of light and is certainly one of the most elegant experiments in physics (because it demonstrates the important concept of interference so simply). For the purposes of this problem, we assume that two long parallel slits extending along the z axis (out of the plane shown in the figure) are separted by a distance . They are illuminated coherently, that is, in phase, by light with a wavelength , for example by a laser beam polarized in the z direction. (Lacking a laser, Young used an intense source diffracted by a slit to produce coherent illumination of his double slits.)

The key point is that the electric field far downstream from the slit (e.g., at a large positive x value) is the sum of the electric fields emanating from each of the two slits. Hence, the relative phase of these electric fields at some


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observation point determines whether they add in phase (constructively) or out or phase (destructively).

To refresh your memory about traveling waves, the electric field that is incident on the double slits from the left is a function of and . Let us assume that it has amplitude . We will also assume a cosine trigonometric function with the arbitrary phase set equal to zero (i.e., at the point you find that ). Then

.

The argument of the cosine function is the phase . It can be written as , where is the angular frequency ( ) and is the wave number defined by . The phase increases by each time the distance increases by . Moreover, the phase is constant for an observer moving in the positive x direction at the speed of light, i.e., for whom .

Part A

Now consider the electric field observed at a point that is far from the two slits, say at a distance from the midpoint of the segment connecting the slits, at an angle from the x axis. Here, far means that , a regime sometimes called Fraunhofer diffraction.

The critical point is that the distances from the slits to point are not equal; hence the waves will be out of phase due to the longer distance traveled by the wave from one slit relative to the other. Calculate the phase of the wave from the lower slit that arrives at point .

Hint A.1 Definition of phaseHint not displayed

Part A.2 Calculate the extra distancePart not displayed

Part A.3 Find the phase of the lower wavePart not displayed

Express your answer in terms of , , , , , , and constants like .




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In order to make the math as simple as possible, we will define two phases:

and

.

Then

and

.

Part CAssuming that the maximum amplitude of the field at point for a wave from midway between the slits is , now find the magnitude of the combined field at due to the two slits. You may ignore variations in the maximum amplitude and consider only variations in phase of the waves emerging from the slits.

Part C.1 Find the magnitude of the electric field due to the lower slitPart not displayed



Part DPart not displayed

Part EPart not displayed

An introduction to thin film interference paired with a good problem

Why Butterfly Wings ShimmerLearning Goal: To understand the concept of thin-film interference and how to apply it.

Thin-film interference is a commonly observed phenomenon. It causes the bright colors in soap bubbles and oil slicks. It also leads to the iridescent colors on many insects and bird feathers. In this problem, you will learn how to work with thin-film interference and see how it creates the dazzling display of a tropical butterfly's wings.

When light is incident on a thin film, some of the light will be reflected at the front surface of the film, and the rest will be transmitted into the film. Some of the transmitted light will be reflected from the back surface of the film. The light reflected from the front surface and the light reflected from the back surface will interfere. Depending upon the thickness of the film, this interference may be constructive or destructive. We will be studying the interference of light normal to the surface of the film. The figure shows the light entering at a small angle to normal only for the purpose of showing the incident and reflected rays.


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For this problem, you will only be concerned with the geometric aspects of thin-film interference, so ignore phase shifts caused by reflection from a medium with higher index of refraction. (Because of the structure of a butterfly's wings, such phase shifts do not contribute much to what you actually see when you look at the butterfly.)

Part AAssume that light is incident normal to the surface of a film of thickness . How much farther does the light reflected from the back surface travel than the light reflected from the front surface?Express your answer in terms of .


Part BFor constructive interference to occur, the difference between the two paths must be an integer multiple of the wavelength of the light (as is true in any interference problem), i.e. the general criterion for constructive interference is

,

where is a positive integer. This is usually stated in the slightly more explicit form

.

Given the thickness of the film, , what is the longest wavelength that can exhibit constructive interference?



Part C

If you have a thin film of thickness 300 , what is the third-longest wavelength of light that exhibits constructive interference with the reflected light?

Note that this corresponds to .

Express your answer in nanometers to three significant figures.


Part D

The criterion for destructive interference is very similar to the criterion for constructive interference. For destructive interference to occur, the difference between the two paths must be some integer number of wavelengths plus half a wavelength:

,

or

,

where is a nonnegative integer. What is the second-longest wavelength that will not be visible (i.e., will have strong destructive interference for the reflected waves) when reflected from a film of thickness 300 ?


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Note that the longest wavelength corresponds to for destructive interference. This is why the notation used for the second-longest wavelength is instead of .

Express your answer in nanometers to three significant figures.


The blue morpho butterfly lives in tropical rainforests and can have a wingspan greater than 15 cm. The brilliant blue color of its wings is a result of thin-film interference. A pigment would not produce such vibrant, pure colors. What cannot be conveyed by a picture is that the colors vary with the viewing angle, which causes the shimmering iridescence of the actual butterfly.

The scales of the butterfly's wings consist of two thin layers of keratin (a transparent substance with index of refraction greater than one), separated by a 200- gap filled with air.

Part EWhat is the longest wavelength of light that will exhibit constructive interference at normal incidence? The keratin layers are thin enough that you can think of them as representing the surfaces of a 200- "film" of air.

Express your answer in nanometers to two significant figures.


Part F

The wavelength that appears in the interference equations given in Parts B and D represents the wavelength of light within the medium of the film. So far we have assumed that the medium composing the film is air, but many thin-film problems will involve films with an index of refraction different from that of air.

Suppose that the butterfly gets wet, thus filling the gaps between the keratin sheets with water ( ). What wavelength in air will be strongly reflected now?

Hint F.1 How to approach the problemHint not displayed

Hint F.2 Relating the wavelength in air to the wavelength in waterHint not displayed

Express your answer in nanometers to two significant figures.


Part GSeveral thin films are stacked together in each butterfly wing scale. How would these multiple layers of thin films affect the light reflected by the butterfly's wings?

Hint G.1 A picture of the situationHint not displayed



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Thin Film (Oil Slick)

A scientist notices that an oil slick floating on water when viewed from above has many different rainbow colors reflecting off of the surface. She aims a spectrometer at a particular spot, and measures the wavelength to be 750 nanometers (in air). The index of refraction of water is .

Part AThe index of refraction of the oil is . What is the minimum thickness of the oil slick at that spot?

Hint A.1 Thin film interferenceIn thin films, there are interference effects because light reflects off the two different surfaces of the film. In this problem, the scientist observes the light that reflects off the air-oil interface, and off of the oil-water interface. Think about the phase difference that is created between these two rays. The phase difference will come from differences in path-length, as well as differences that are introduced by certain types of reflection. Recall that if the phase difference between two waves is (a full wavelength) then the waves interfere constructively, while if the phase difference is (half of a wavelength) the waves interfere destructively.

Hint A.2 Path-length phase differenceThe light that reflects off the oil-water interface has to pass through the oil slick, where it will have a different wavelength. The total "extra" distance it travels is twice the thickness of the slick (once as it moves toward the oil-water interface, and once as it reflects back, out into the air).

Hint A.3 Phase shift due to reflectionsRecall that when light reflects off a surface with a higher index of refraction, it gains an extra phase shift of radians. What used to be a maximum is now a minimum! Be careful, though; if two beams each reflect off a surface with a higher index of refraction, they will both get a phase shift, canceling out that effect.

Express your answer in nanometers, to three significant figures.

ANSWER: = 313

Part BSuppose the oil had an index of refraction of 1.50. What would the minimum thickness be now?

Hint B.1 Phase shift due to reflectionsKeep in mind that when light reflects off a surface with a higher index of refraction, it gains an extra phase shift. What used to be a maximum is now a minimum! Be careful, though; if two beams reflect, they will both get a phase shift, canceling out that effect. Also, reflection off a surface with a lower index of refraction yields no reflection phase shift.


ANSWER: = 125

Part CNow assume that the oil had a thickness of 200 and an index of refraction of . A diver swimming underneath the oil slick is looking at the same spot as the scientist with the spectromenter. What is the longest wavelength

of the light in water that is transmitted most easily to the diver?

Hint C.1 How to approach this partFor transmission of light, the same rules hold as before, only now one beam travels straight through the oil slick and into the water, while the other beam reflects twice; once off the oil-water interface, once again off of the oil-air interface, before being finally transmitted to the water.

Part C.2 Wavelength of light in air


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Find the wavelength of the required light, in air.Express your answer numerically in nanometers.


Hint C.3 The wavelength of light in waterHint not displayed


ANSWER: = 451

This problem can also be approached by finding the wavelength with the minimum reflection. Conservation of energy ensures that maximum transmission and minimum reflection occur at the same time (i.e., if the energy did not reflect, then it must have have been transmitted in order to conserve energy), so finding the wavelength of minimum reflection must give the same answer as finding the wavelength of maximum transmission. In some cases, working the problem one of these ways may be substantially easier than the other, so you should keep both approaches in mind.

Using an interferometer

Using a Michelson InterferometerYou are asked to find the index of refraction for an unknown fluid, using only a laser and a Michelson interferometer. A Michelson interferometer consists of two arms--paths that light travels down, which end in mirrors-- attached around a beam splitter. The beam splitter separates the incoming light into two separate beams and then recombines them once they return from the ends of the arms. The recombined beams are sent to a telescope, where their interference pattern may be observed in detail.

Part AFirst, you must find the wavelength of the laser. You shine the laser into the interferometer and then move one of the mirrors until you have counted fringes passing the crosshairs of the telescope. The extremely accurate micrometer shows that you have moved the mirror by millimeters. What is the wavelength of the laser?

Hint A.1 Relating wavelength and distance in a Michelson interferometerHint not displayed

Express your answer in nanometers, to four significant figures.



Summary 2 of 10 problems complete (19.74% avg. score)9.87 of 10 points

a simple introduction to interference - university …jbs2/edlish2/heimilishrrr/35.pdfa simple...

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