In this chapter the "wave-like behavior" of light will be studied. There is a simple experiment that verifies light is a wave. The experiment is called the "Young's Double-slit experiment." To learn about Young's experiment, we need to have an understanding of "constructive and destructive interferences."
Constructive and Destructive Interference of Waves:
Two waves with the same wavelength and frequency arriving simultaneously at a point in space or a medium can have an increased effect at that point or to some extent neutralizing effect. The two arriving waves interfere with each other. If they happen to be in phase at that point, the overall effect will be equal to the sum of the individual effects. This interference is called "constructive interference." If they are out of phase at that point, the overall effect depends on how much out of phase one wave is relative to the other. The maximum neutralizing effect occurs when they are 180o out of phase that results in the greatest "destructive interference."
Figures 1 and 2 below, show two extreme scenarios: 1) the two waves are completely in Phase and 2) they are completely out of phase.
When completely in phase (Fig. 1), their amplitudes add up, and the resulting wave has an amplitude equal to A1 + A2.
When completely out of phase (Fig. 2), their amplitudes subtract, and the resulting wave has an amplitude equal to |A1 - A2|. If the two180o-out-of-phase interfering waves have also equal amplitudes at that point, they result in a vanished effect at that point. You have this experience with your cellular phone. Near buildings, your phone set can receive waves not only from a local cellular antenna, but also the reflection of the same wave from a nearby building. If you experience "call fading," at a certain instant, it is because of the fact that the two waves coming to your cell phone are completely out of phase at where you are, at that instant.
In Fig. 2, if the two waves have the same amplitude, i.e., A1 = A2, then they cancel each other and A = 0.
Two or more sources are called coherent if not only they have the same frequency and the same wavelength, but also they are in phase.
Young's Double-Slit Experiment:
Suppose you have a small rectangular box one side of which is made of a thin sheet of metal with two parallel and very closely spaced slits cut in it. Also, suppose that there is a bright light source (a point source) inside the box such that light can emerge from the box through those two slits only. We further suppose that the light source is located on the perpendicular bisector of the two slits. This means that the light source is equidistant from the slits. If you are in a dark room and the only way light can come out of the box is via the two slits, you expect to see two bright lines on the wall parallel to the slits. To your surprise, you will see several bright light lines on the wall! The only way this phenomenon can be explained is that light is a wave. Young was the first to explain this phenomenon.
The slits, (being equidistant from the light source in the box), receive light waves from a single source and become coherent sources. Each slit becomes an independent source. These two independent and coherent sources send out waves in all directions. To the right of the sources (slits), in Fig. 3, there are many points at which the waves coming from these two sources are in phase and add up constructively. There are also many point at which the arriving waves interfere destructively. Wherever they combine constructively, they form a bright streak of lightthat Young called it a "bright fringe." Wherever they combine destructively, they form a dark streak that he called it a "dark fringe." The fringes pattern is shown below:
Bo is where the central bright fringe is formed. As shown in Fig. 3, at Bo , light waves coming from S1 and S2 are in phase and interfere constructively to form a bright fringe. The reason for being in phase at Bo is that S1Bo = S2Bo. The distance difference is zero or (0λ). Each of S1Bo & S2Bo accommodate the same number of wavelengths (λ's) in it. Since the waves that leave S1 and S2 are in phase, they will arrive at Bo, in phase as well.
At D1 above Bo, a dark fringe is formed. Light coming from S2 travels a longer distance than the light coming from S1. The distance difference must be (1/2)λ that makes the two waves 180o out of phase there.
At B1 above D1, a bright fringe is formed. Light coming from S2 travels a longer distance than light coming from S1. The distance difference must be exactly 2(λ/2) that makes the two waves in phase there.
At D2 above B1, a dark fringe is formed. Light coming from S2 travels a longer distance than the light coming from S1. The distance difference must be 3(λ/2) that makes the two waves 180o out of phase there.
At B2 above D2, a bright fringe is formed. Light coming from S2 travels a longer distance than light coming from S1. The distance difference must be exactly 4(λ/2) that makes the two waves in phase there.
The above repeating arguments can go on for as many bright and dark fringes as are made. The same argument is also true for the fringes below Bo.
Conclusion: From the above reasoning, one may conclude that it is the distance difference that is important and that
1) for a bright fringe to form, the distance difference must be an even multiple of λ/2, and
2) for a dark fringe to form, the distance difference must be an odd multiple of λ/2.
Note that even multiples of λ/2 are: 0(λ/2), 2(λ/2), 4(λ/2), 6(λ/2), ... or, simply 0, 1λ, 2λ, 3λ, ... or, mathematically nλ wheren = 0, 1, 2, 3, ... .
Also, odd multiples of λ/2 are: 1(λ/2), 3(λ/2), 5(λ/2), 7(λ/2), or, mathematically (2n - 1)λ/2 where n = 1, 2, 3, 4, ... .
In practice, S1 and S2 are made very close together, often much closer than one tenth of a millimeter. The distance between the two coherentsources is shown by d. The screen is generally placed at least 1m from the slits. The two rays emerging from the two slits are very closely packed. The angles that, for example, S2B1 and S1B1 make with the central line are practically equal. For the first bright fringe, the angle is labeled θ1 . The angle that nearly both S2B2 and S1B2 make with the central line is labeled θ2. See Fig, 4. The fringes below the central line are not shown due to symmetry.
Young's Formula for the angle of the n-th Bright Fringe:
It is easy to derive a formula for the angle corresponding to the n-th bright fringe. The formula relates θn , d, and λ , the wavelength of the light used. The formula is:
where n = 0, 1, 2, 3, ... . In the proof and in Fig. 5, note that the two angles that are labeled θn are equal because their sides are perpendicular to each other.Let's magnify the circled area in the above figure only for the n-th bright fringe for which the distance difference is S2 A =S2 Bn - S1Bn = nλ.
In the right triangle S1S2A, the sine of angle θn is: sinθn = S2A / S1S2 ,or
sinθn = nλ /d for bright fringes as expected.
|Using Bn ,
the n-th bright
fringe on the
wall, as the center of the circle for which S1Bn is
the radius, arc S1Acan
be drawn to find the distance
difference S2 Abetween S1Bn and S2Bn.
Note that S1Bn and S2Bn are so close that arc S1A becomes perpendicular to both of them making S1S2A a right triangle.
Example 1: The 3rd bright fringe in a double slit experiment makes a 2.4o angle with respect to the central line. The wavelength of the monochromatic light used is 480nm. Find the distance d between the two sources.
Solution: sinθn = nλ/d ; d = nλ /sin θn ; d = 3(480x10-9)/sin2.4o.
d = 34.4x10-6m = 0.0344x10-3m = 0.0344mm.
Young's Formula for the angle of the n-th Dark Fringe:
| It is easy to use a similar
argument as above to show that the angles corresponding to dark
fringes in a
double-slit experiment may be found from the formula on
the right where n = 1,
2, 3, ... .
There is no such thing as central dark fringe. n = 1 corresponds to the 1st dark fringe on each side of the central bright.
Note that (n - 1/2)λ is the same as (2n -1)λ/2 that means odd multiples of λ/2.
Example 2: Find the angle corresponding to the 5th dark fringe with respect to the centerline in a double slit experiment knowing that the wavelength used is 550nm and the sources are 0.022mm apart.
Solution: sinθn= (n - 1/2)λ/d ; sinθ5 = (5 - 1/2)550x10-9m/0.022x10-3m.
θ5 = 6.5o.
Alternate Method to Measure Angle θn :
As shown in Fig. 6: D, y1, y2, y3, and so on can be measured. From such measurements θn , the angle corresponding to that n-th fringe can then be calculated using a tangent ratio as shown below:
Example 3: The distance between the 4th bright fringe to the central fringe (on the screen) in a double-slit experiment is 23.0mm and the slits are 7.520m from the screen. If the distance between the slits is 0.560mm, determine the wavelength and frequency of the light used.
Solution: tanθn = yn/D ; tanθ4 = 23.0/7520 ; θ4 = 0.175o.
sinθ4= 4λ/d ; sin0.175o = 4λ/0.560x10-3m ; λ = 428x10-9m = 428nm.
c = f λ ; f = c /λ ; f = (3.00x108m/s)/428x10-9m = 7.00x1014Hz.
Example 4: If in a double-slit experiment, the slits spacing is 5.00x10-6m, and light of wavelength 655nm is used, how many bright fringes do form on each side of the central bright fringe?
Solution: sinθn= nλ/d ; sinθn= n(655x10-9m)/5.00x10-6m ; sinθn= 0.131n.
Giving n different values, we may determine the number of possible fringes as in the following chart:
As can be seen, giving different values to n insinθn = 0.131n, results in 7 possible angles at which bright fringes occur.
For n = 8, we don't get a fringe, because the sine of an angle can not be more than 1.
An easier way to do this is to set sinθn = 1. This means setting 0.131n = 1 from which we get:
n = 1/0.131 = 7.64. Rounding to the lower integer, we get: n = 7.
|0||0||0 (Central Bright)|
|8||1.048 > 1||Not Possible|
Test Yourself 1:
1) Two waves completely cancel each others effects if they have the same amplitude and are (a) 90o out of phase (b) 180o out of phase (c) 360o out of phase. click here.
2) Constructive interference occurs when two waves are (a) in phase (b) completely out of phase (c) neither a nor b.
3) Two waves are in phase if they (a) reach maximum together (b) reach minimum together (c) become zero together (d) a, b, & c. click here.
4) When two waves that are completely in phase at a point interfere, the amplitude of the resultant wave is (a) the product A1A2 (b) the sum A1 + A2 (c) neither a nor b.
5) For two waves to be coherent, they must (a) be in phase (b) have the same wavelength (c) have the same frequency (d) a, b, &c. click here.
6) A light source S that is equidistant from two slits makes the two slits like two independent sources that (a) are in phase (b) have the same wavelength (c) the same frequency (d) the same amplitude (e) a, b, c, & d.
7) A light source S that is equidistant from two slits makes the two slits like two independent sources that are called coherent. (a) True (b) False click here.
8) In Young's Double-slit experiment, we may say that any point in space that is equidistant from the slits will be at constructive interference. (a) True (b) False
9) In Young's Double-slit experiment, for a point on the screen to be at constructive interference, it must have a distance difference from the two slits of (a) 0λ b)1λ c) 2λ d) 3λ e) 4λ. ...... f) nλ where n = 0, 1, 2, ....
10) In Young's Double-slit experiment, for a point on the screen to be at destructive interference, it must have a distance difference from the two slits of a)1λ/2 b) 3λ/2 c) 5λ/2 d) 7λ/2 ...... e) (n-1/2)λ, where n = 1, 2, 3, ....
11) The distance between the slits d in the Young's experiment can easily be (a) 3000 times (b) 5000 times (c) 10,000 times (d) a, b, & c smaller than D the distance between the slits and the screen. click here.
12) Connecting the slits to a point on a screen by two lines makes the lines (a) close to each other (b) far from each other (c) so close to each other that they appear as one line. click here.
13) Since the lines that connect the slits to a fringe on the screen are almost on each other, we may treat it as one line and name the angle it makes with the central line as θn , the angle corresponding to the n-th bright or dark fringe. (a) True (b) False
14) The angle θn corresponding to the n-th bright fringe may be found from (a) sinθn = nλ (b) dsinθn= nλ (c) a & b. Of course, n =0, 1, 2, 3, ... . click here.
15) The angle θn corresponding to the n-th dark fringe is (a) dsinθn = (n-1/2)λ (b)dsinθn = nλ (c) a & b. Of course n = 1, 2, 3, ... .
16) θn may also be calculated from the equation tanθn = yn /D2 where yn is the distance from the central bright fringe to the n-th bright or dark fringe on the screen. (a) True (b) False click here.
Problem: In a Young's experiment, light of wavelength 482nm is used and the angle to the 4th bright fringe is measured 0.50o with respect to the central line. The distance from the 4th bright fringe to the central bright is 17mm. Answer the following questions:
17) The distance between the slits is (a) 0.22cm (b) 0.22mm (c) 0.0040mm.
18) The distance between the slits and the screen is (a) 1.95m (b) 6.4ft (c) a & b. click here.
In a Young's experiment, the distance between the slits is 0.123mm, and the screen is 2.40m away from the slits. Answer the following questions:
19) If the distance from the 3rd dark fringe to the central bright is 31.7mm, the wavelength of the light used is (a) 650nm (b) 540nm (c) 410nm. click here.
20) The distance from the 5th bright fringe to the central one is (a) 40.3mm (b) 63.4mm (c) 35.0mm.
A Brief on The Reflection of Mechanical Waves:
Recall that when a mechanical wave hits a hard medium (like waves in a stretched string that travel back and forth in between two hard fixed ends), the reflected wave goes 180 o or λ /2 out of phase relative to the incident wave.
On the other hand, when a mechanical wave hits a soft medium, both of the reflected and refracted waves remain in phase relative to theincident wave.
Next, in the analysis of thin film interference, the same is true for light waves.
When light waves hit a higher refraction index medium, the reflected waves go completely (λ /2) out of phase with respect to the incident waves.
When light waves hit a lower refraction index medium, the reflected and refracted waves stay in phase with respect to the incident waves.
reflected ray may
or not be in phase with the incident ray depending on
refraction indices; however,
2) a refracted ray is always in phase with the incident ray.
A thin film of gasoline or soap on water creates colorful patterns that we all have noticed. As shown in Fig. 7, the phenomenon is caused by the interference of the immediately reflected rays at A and those reflected at B.
The goal is to determine if Rays 1 and 2 that go to the observer's eyes are in phase or out of phase. There are two deciding factors at work here:
1) the refraction indices of air, gasoline, and water
2) the thickness of the gasoline layer itself.
refraction index effect:
Ray IA hits the gasoline surface (higher n than air) and thereflected Ray 1 is 180o or λ /2 out of phase relative to IA.
AB, the refracted ray, is in phase with IA. BC is reflected at B at a lower n of water compared to gasoline. BC remains in phase with AB and therefore in phase with IA. C2 is also a refracted ray and is in phase with BC, AB, and IA.
Result: Ray 1 is λ /2 out of phase with IA, but Ray 2 is in Phase with IA; thus, Rays 1 and 2 are out of phase by λ /2and the observer sees the gasoline surface to be dark. This conclusion could be wrong!
The Reflection at A causes λ /2 phase difference, but reflection at B that is at an optically softer medium occurs without a phase difference.
2) The Film Thickness Effect: What we discussed under 1 is only half of the story! The thickness of the film itself is another deciding factor as to whether Rays 1 and 2 will be in phase or not. We will examine a few thicknesses to get the idea.
i) If thickness t of the film is λ /4. AB + BC accommodate 2(λ /4) = λ /2. This means that the AB+BC trajectory itself causes another λ /2 phase difference and Rays 1 and 2 become in phase as seen by the observer. The observer sees the gasoline surface to be bright!
ii) If thickness t of the film is λ /2. AB + BC accommodate 2(λ /2) = λ. This means that the AB+BC trajectory does not cause any phase difference and Rays 1 and 2 will still be out of phase. The observer sees the gasoline surface to be dark.
iii) If thickness t of the film is 3λ/4. AB + BC accommodate 2(3λ /4) = 3λ /2. This means that the AB+BC trajectory itself causes another λ /2 phase difference and Rays 1 and 2 become in phase as seen by the observer. The observer sees the gasoline surface to be bright and so on... .
It is now clear that if the film thickness t is an odd multiple of λ /4, both exit rays will be in phase and the observer sees the gasoline surface bright; therefore, the condition to have bright fringes is:
t = (2k-1)λ /4 where k = 1, 2, 3 ...
|Note that λ is the wavelength of light in the film's medium (material).|
The Wavelength and Refraction Index:
Example 5: The refraction index of air is 1.00 and that of gasoline is 1.40. The wavelength of a certain green light in air is 532nm. Find its wavelength in gasoline.
Solution: Using n1λ1 = n2λ2 ; 1.00(532nm) = 1.40λ2 ; λ2 = 380nm.
In brief, the reflected Ray 1 is 180o out of phase with respect to IA. The Reflected Ray 2 is also somewhat out of phase with respect toIA. This means that in most cases reflected Rays 1 and 2 are somewhat out of phase. The degree of being out of phase can vary from zero to 180o depending on the angle of incidence and the thickness of the film. We should not forget that white light is also a mixture of so many different colors (wavelengths) for which angles of refraction differ and therefore the amount that each color is out of phase is also different.The combination of a large number of reflections, refractions, and dispersions give rise to such beautiful patterns on soap or gasoline films. It sometimes happens that for a certain thickness of the gasoline layer and angle of incidence a certain color is more pronounced or completely eliminated. That is when the reflected Rays 1 and 2 are either completely in phase or completely out of phase.
Example 6: To an observer looking at an angle that is almost straight down, the color of a thin film of gasoline appears bright yellow. The wavelength of yellow light in air is 580nm. What are the first three possible thicknesses (t's) of the gasoline film for this to happen?
Solution: The assumption is that the incident rays are at small angles (close to their respective normal lines). The wavelength of yellow light in gasoline is
n1λ1 = n2λ2 ; 1.00(580nm) = 1.40λgas. ; λgas. = 414nm.
The observer sees bright yellow light. It means that both reflected Rays 1&2 are in phase and interfere constructively. The condition t = (2k-1)λ /4 must be met. We may write:
t = (2k-1)λ /4
where k = 1, 2, 3 ...
|Case||k||t = (2k-1)λ /4|
|1||1||t = 1λ /4 = 1(414nm) /4 = 104nm|
|2||2||t = 3λ /4 = 3(414nm) /4 = 310nm|
|3||3||t = 5λ /4 = 5(414nm) /4 = 518nm|
|An air wedge is the space between two sheets of glass (or transparent media) that make a small angle with each other. As Fig. 9 shows, the actual air wedge is limited by the two solid lines. At the top line, light already in glass hits the glass-air interface for which the reflected ray 1 at A is in phase with the incident ray ( nglass > n air). At the bottom edge of the wedge, light already in air hits the air-glass interface and the reflected ray 2 at B is λ /2 out of phase with respect to the incident ray AB ( n air < nglass ). In the air wedge, the ray travels two extra distances, one from A to B, and the other from B to C. This travel distance of 2t creates some phase difference between the final Rays 1 and 2.||
The combination of the the λ/2 that occurs at B and the partial phase difference created by 2t give rise to possible constructive and destructive interference between typical rays1 and 2.
Since the air gap varies linearly for different parallel incident rays, both constructive and destructive interferences occur for some of the incident rays. The result is the formation of bright and dark fringes as viewed by observers looking nearly straight down into the wedge. The condition for bright fringes (constructive interference) is:
2t + λ/2 = kλ where k = 1, 2, 3, ... or, 2t + λ/2 = 1λ, 2λ, 3λ, ...
Example 7: Show that the above result of 2t + λ/2 = kλ is the same as the previous result of t = (2k - 1)λ/4.
Solution: To be done by students.
Diffraction is the bending of light upon passing through small openings or by sharp edges. Diffraction occurs in ocean waves as well. Actually, it is beneficial to first examine the diffraction concept with ocean (machanical) waves arriving at a jetty. One role of a jetty where boats are parked is to separate the body of the water in the jetty from the heavy waves coming in from the ocean. That way the parked boats receive much weaker (in amplitude) waves and do not bump into each other. Each jetty has a small opening to the ocean compared to its body of water. See Fig. 10. The opening is made large enough to allow all size boats to comfortably pass through ; however, its size is much smaller than the jetty itself.
As shown in Fig. 10, the wavefronts arriving from the ocean hit the jetty like almost straight lines one after each other. See Remark 1 in Fig. 10. The huge rocks bordering the jetty break the arriving fronts except at the opening. The straight fronts make the lump of water at the jetty's opening oscillate up and down accordingly. The good size lump of water at the opening becomes an independent source that sends waves in all directions especially into the jetty. That lump of water being the originator of new waves is at the center of this wave generator. The new waves center is the jetty's opening itself. The new wavefronts are not straight fronts anymore. The are very curved. See Remark 2 in Fig. 10. If they are given a huge area to propagate in, they become more straight similar to the ones that keep coming from the ocean. The new waves into the jetty propagate in different directions compared to the waves arriving from the ocean side. For an observer looking down from a helicopter into the jetty, it appears that the ocean waves bend when they pass through the opening. The observer will call this phenomenon the "diffraction" of the ocean waves. The same concept applies to light waves as they pass through small openings or by sharp edges.
When light waves arrive at an opening or an aperture, each point of the opening becomes an independent source sending wavelets in all directions. The wavelets emerging from such independent sources interfere at different points of the region past the aperture causing interference patterns due to diffraction on a wall or a screen (Fig. 11). If the aperture or the hole that a wavefront arrives at is big compared to the wavelength of the light, the diffraction pattern will not be very clear and approximates the shape of the aperture (hole) itself; however, if the opening's diameter is relatively small (as much as a number of wavelengths), sharper diffraction patterns are formed.
The diffraction pattern (on the wall) of light passed through a small aperture.
When either the source or the screen is near an aperture or an obstruction (a sharp edge), the wavefronts are spherical (not straight) and the diffraction pattern they form is quite complex. This is called "Fresnel Diffraction." One case is shown in Fig. 12.
In this case some light enters the region of geometrical shadow.
The curve shows how the intensity and width of the bright fringes change.
Fraunhofer diffraction is one in which source is far from the slit or aperture. The wavefronts arriving at the slit or aperture are almost like straight lines. In single-slit diffraction, the diffraction pattern on the screen depends on the aperture diameter or width, α. Note that a single slit has a width only and not a diameter. If the aperture diameter or the single slit width is large (several times the wavelength λ, the lit area over the screen fairly defines the shape of the aperture or the slit. If the aperture diameter or the slit width is small and comparable to the wavelength of the light used, the diffraction patterns will be more pronounced or have better contrast as shown in Fig. 13.
Interference of N Coherent Sources
When a large number of coherent sources interfere, interesting interference patterns form on a screen parallel to the sources. A diffraction grating is a thin clear film that some 5000 lines per inch (some 200 lines per mm) is scraped on it by a special technique. See Fig. 14. If a narrow laser beam of 1mm in diameter is incident on such grating (normal to it), the laser beam will be divided into 200 much narrower beams. In other word, it is like having 200 coherent sources that each send light in all directions past the film (grating) to interfere. That is the trick to make N coherent sources so close to each other. The space in between every two neighboring lines on the grating that allows laser waves to pass through becomes an independent source. The interference patterns of these coherent sources can then be observed on the wall (screen). The diffraction pattern is comprised of a number of principal maxima (bright fringes) with many more weaker maxima in between.
The weaker maxima in between the principal maxima are very difficult to see. Those are very weak in intensity. The mathematics for this study is more involved. Here we just suffice with showing the patterns and introducing the formula. The formula that calculates the angles corresponding to the n-th principal maxima is similar to the one for Young's double-slit experiment. The N coherent sources and the relative intensity of their principal maxima are shown below:The formula that calculates the angles corresponding to the principal maxima is
where n = 0, 1, 2, 3, ... ...correspond to the central, 1st, 2nd, 3rd, and ... principal maxima. In most cases, the central, 1st, 2nd, and 3rd principal fringes are bright enough to be seen, the rest are normally dim.
As Fig. 14 shows, the central maxima is very bright. The intensities of the first and second maxima on each side drop sharply. Suppose only 202 lines are involved in making only 201 coherent sources. All 201 sources will have a contribution to the central maximum. The reason is symmetry. The very middle source is the 101-th source. The beam from this source will be exactly along the centerline. Sources 100 and 102 on the grating will be equidistant from the center point on the wall and will interfere constructively. Same is true for pairs (99 &103), (98 &104), (97 &105), and so on. This means that all 201 sources are in phase (interfering constructively) at the center on the wall. That is why and how the central principal maxima is the brightest. There is no guarantee that all sources will be in phase at the 1st, 2nd, or 3rd principal maximum on either side of the central one. In fact a much less number of sources are in phase at the 1st principal maximum and a much much less number of sources are in phase at the 2nd principal maximum. For 10,000 coherent source, all 10,000 are in phase at the center (n = 0), 250 are in phase at the 1st maximum (n = 1), and only 16 are in phase at the 2nd maximum (n = 2). That's why the intensity of light drops so sharply for subsequent principal maxima.
Example 7: A diffraction grating is positioned 120.0cm from a wall and forms interference fringes (pattern) as a 635nm laser beam passes through it. The very closely-spaced lines on the grating create many independent and coherent sources at the film as shown in the above figure. The distance between the second principal maxima to the central one measures 31.5cm. Find the number of lines per millimeter of the film (grating).
Solution: tanθn= yn/D ; tanθ2= 31.5cm/120.0cm ; θ2=14.71o ; sinθ2= 2λ/d.
sin14.71o = 2(635x10-9m)/d ; d = 5.00 x10-6m/line = 5.00x10-3mm/line.
# of lines per mm = 1/d. ; N = 1/(5.00x10-3mm/line) = 200 lines/mm.
Chapter 37 Test Yourself 2:
1) Thin film interference occurs when (a) light passes through a thin plastic film (b) light passes through a photographic film (c) a thin layer of oil or gasoline is on another liquid like water. click here.
2) The reason for thin film colorful patterns is (a) the oiliness of the oil or gasoline on water (b) the image of colorful objects reflected from the oil or gasoline layer (c) the interference of the reflected rays from the gasoline surface and the underlying water surface.
Refer to Fig. 7 and answer the following questions: click here.
3) At Point A, the refracted ray AB is (a) in phase with the incident ray IA (b) completely out of phase with the incident ray IA (c) partially out of phase with the incident ray IA.
4) The first reflected ray A1 is (a) in phase with the incident ray IA (b) completely out of phase with the incident ray IA (c) partially out of phase with the incident ray IA. click here.
5) The reasoning for Question 4 is that (a) gasoline (n = 1.4) is optically harder than air (n = 1.00) and the reflected ray A1 becomes completely out of phase (b) gasoline is optically softer than air, and the reflected ray A1 becomes completely out of phase (c) neither a nor b.
6) At B, the reflected ray BC is (a) in phase with the incident ray AB (b) partially out of phase with the incident ray AB (c) completely out of phase with the incident ray AB. click here.
7) The reasoning for Question 6 is that (a) water (n = 1.00) is optically harder than gasoline (n= 1.4) and the reflected ray BC becomes completely out of phase (b) water (n = 1.00) is optically softer than gasoline (n = 1.4) and the reflected ray BC stays in phase with ray AB (c) neither a nor b.
8) At C, the refracted ray C2 is (a) in phase with BC. (b) completely out of phase with BC. (c) partially out of phase with BC. click here.
9) If the thickness of the gasoline layer is such that it does not cause a phase difference, then rays IA and C2 will be (a) completely out of phase (b) partially out of phase (c) in phase.
10) If the thickness of the gasoline layer is such that it does not cause a phase difference, then rays A1 and C2 will be (a) completely out of phase (b) partially out of phase (c) in phase. click here.
11) If the thickness of the gasoline layer is such that it does not cause a phase difference, then rays A1 and C2 will be completely out of phase and destructive interference occurs in which case the gasoline surface appears dark for a person who is looking almost straight down onto it. (a) True (b) False
12) The wavelength of the light ray in the gasoline layer is (a) shorter than it is in air (b) longer than it is in air (c) same as it is in air. click here.
13) Referring to Question 11, in most cases however, the thickness of gasoline is such that the overall phase difference between rays A1 andC2 is not exactly 180o, and interference can occur in a wide range. (a) True (b) False
14) It is the result of a wide range of interference between the reflected rays along with dispersion through refraction that the beautiful and colorful patterns form. (a) True (b) False
Redraw the ray diagram in Fig. 7 and repeat answering the above questions again. click here.
Air Wedge: Refer to Fig. 9 and answer the following questions: click here.
15) The first incident ray at I is (a) in phase with IA (b) 180o out of phase with IA (c) partially out of phase with IA.
16) The reflection at A is (a) in phase with IA (b) 180o out of phase with IA (c) partially out of phase with IA.
17) The refracted ray 1 is (a) in phase with IA (b) 180o out of phase with IA (c) partially out of phase with IA.
18) Overall, Ray 1 is (a) in phase with the original incident ray (b) 180o out of phase with the original incident ray (c) partially out of phase with the original incident ray. click here.
19) At A, part of ray IA refracts and becomes AB that enters air. The refracted ray AB that has entered an optically softer medium is (a) in phase with IA (b) 180o out of phase with IA (c) partially out of phase with IA.
20) At B, the reflected ray BC is (a) in phase with AB (b) 180o out of phase with AB (c) partially out of phase with AB.
21) Rays CD and D2 are (a) in phase with AB (b) 180o out of phase with AB (c) partially out of phase with AB.
22) Rays CD and D2 are (a) in phase with the original incident ray (b) 180o out of phase with the original incident ray (c) partially out of phase with original incident ray. click here.
23) Overall, if the air gap does not contribute to a phase difference while the ray travels in it, Ray1 and Ray2 are (a) in phase (b) 180o out of phase (c) partially out of phase.
24) Since the air gap thickness varies linearly, it contributes to a varying phase difference from each point to the next. This causes the final outgoing rays to be completely in phase and completely out of phase alternatively. This result in an interference pattern and several dark and bright fringes can seen by an observer that is looking almost straight down onto the glass (the air wedge). (a) True (b) False click here.
25) Diffraction is (a) the bending of light when it changes medium (b) the separation of light into its constituent colors upon refraction through a prism (c) the bending of light as it passes through small openings or by sharp edges. click here.
26) A diffraction grating is (a) a thin sheet of transparent material that has a large number of closely-spaced lines per inch of it (b) very helpful in measuring the wavelength of visible light (c) a & b.
27) A diffraction grating may be viewed as a device that generates a large number of coherent sources. (a) True (b) False
28) The coherent sources produced by a diffraction grating are not only in phase, but also have the same (a) wavelength (b) frequency (c) amplitude (d) a, b, & c. click here.
29) The formula that determines the angle of the n-th principal maximum with respect to the central one that a diffraction grating forms is (a) sin θn = nλ (b) sin θn = λ/d (c) sin θn= nλ/d.
30) In the formula sinθn = nλ/d, or dsinθn = nλ, d is (a) the width of the grating (b) the space between every two adjacent lines of the grating (c) both a & b. click here.
31) If N is the number of lines per mm of a diffraction grating, then (a) d = 1/N (b) d is the number of mm/line. (c) both a & b.
Problem: A diffraction grating forms the interference pattern of a laser beam (λ = 654nm) on a wall 121cm from it. The distance from the 2nd principal maximum to the central one is 28.3cm. Answer the following questions: click here.
32) The angle that the 2nd principal maximum makes with respect to the central line is (a) 1.5o (b) 13.2o (c) 23.2o.
33) The distance between every two adjacent slits, d, of the grating is (a) 5.72x10-6m (a) 0.00572mm (c) both a & b.
34) The number of lines per mm of the grating is (a) 175 (b) 200 (c) 208. click here.
Problem: A diffraction grating has 5080 lines/inch. At a distance of 244cm from a wall, a student is looking through this grating at a lit bulb on the wall. The bulb emits bright white light. The grating forms rainbow spectra on both sides of the bulb on the wall as seen by the student. You may ask your teacher to set up this simple experiment in class, if possible. You may also be given a grating to experiment on your own. The student and his/her classmate measure the red and blue ends of the first principal maximum to the center point where the bulb is as (Y1) red = 33.9cm and (Y1)blue = 20.1cm. Answer the following questions:
35) The number of lines per mm of the grating is (a) 100 (b) 175 (c) 200. click here.
36) The number of mm per line or the "d" of the grating is (a) 0.0100mm (b) 0.00571mm (c) 0.00500mm.
37) d expressed in meters per line is (a) 1.00x10-5m (b) 5.71x10-6m (c) 5.00x10-6m. click here.
38) As seen by the student, using the (tan-1 function), the angle that the blue end makes with the central line (the line from the student's eye to the bulb) is (a) 4.71o (b) 5.71o (c) 6.71o.
39) Using dsinθn = nλ for principal maxima of the grating's pattern, the wavelength of blue light is (a) 504nm (b) 411nm (c) 560nm. click here.
40) As seen by the student, using the (tan-1 function), the angle that the red end makes with the central line (the line from the student's eye to the bulb) is (a) 5.63o (b) 7.91o (c) 8.51o.
41) Using dsinθn = nλ for principal maxima of the grating's pattern, the wavelength of red light is (a) 404nm (b) 511nm (c) 688nm. click here.
1) The 5th bright fringe in a double slit experiment makes an 9.50o angle with respect to the centerline. The wavelength of the monochromatic light used is 532nm. Find the distance between the two neighboring sources.
2) Find the angle for the 7th dark fringe in a double slit experiment with respect to the centerline knowing that the wavelength used is 650nm and the sources are 0.011mm apart.
3) The distance between the 3rd bright fringe to the central fringe (on the screen) in a double-slit experiment is 42mm and the slits are 840mm from the screen. If the distance between the slits is 0.0289mm, determine the wavelength and frequency of the light used.
4) If in a double-slit experiment, the slits spacing is 0.0080mm, and light of wavelength 635nm is used, how many bright fringes do form on the screen?
5) To an observer looking almost straight down, the color of a thin film of gasoline appears bright green. The wavelength of green light in vacuum is 532nm. What are the first three possible thicknesses (t's ) of the gasoline film for this to happen? The refraction index of gasoline is 1.40.
6) A diffraction grating positioned at 133.0cm from a wall forms interference fringes on the wall as a 532nm laser beam passes through it. The distance between the 3rd principal maximum to the central one measures 44.8cm. Find (a) the distance between every two neighboring lines on the grating, and (b) the number of lines per millimeter of it.
1) 0.0161mm 2) 22.6o 3) 481nm 4) 25
5) 95nm, 285nm, 475nm 6) 5.00x10-6m, 200. lines/mm,