In this chapter wavelike 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." Before this experiment is explained, we need to understand what constructive and destructive interference of two waves mean.
Constructive and Destructive Interference of Waves:
When two waves of the same wavelength that are in phase arrive at the same point, they completely amplify each other and the resulting wave has an amplitude that is equal to the sum of the amplitudes of the arriving waves as shown in Fig.1.
When two waves of the same wavelength that are 180° out of phase arrive at the same point, they act opposite to each other and the resulting wave has an amplitude that is equal to the difference of the amplitudes of the arriving waves as shown in Fig.2. 180° out of phase means when one is at its maximum, the other is at its minimum.
In Fig. 2, when the two waves have the same amplitude, i.e., A1 = A2, then the two waves cancel each other and A = 0.
Two or more sources are called coherent if they have the same frequency, the same wavelength, and 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 the 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 these two slits, you expect to see two bright lines on the wall parallel to the slits. To your surprise, you will see several bright lines form on the wall! The only way this phenomenon can be explained is that light is a wave. Young was the first to explain the 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, in the figure shown, 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 combine destructively. Wherever they combine constructively, they form a bright streak of light that Young called a "bright fringe." Wherever they combine destructively, they form a dark streak that he called a "dark fringe." The fringe patterns, specially on a screen is shown below:
Bo is where the central bright fringe is formed. At Bo , light waves coming from S1 and S2 are in phase and interfere constructively to create a maximum intensity light streak or a bright fringe.
At D1 above Bo, a dark fringe is formed. Light coming from S2 travels a longer distance than the light coming from S1. The difference in the distance must be (1/2)λ that has made the two waves 180° out of phase.
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 (2/2)λ or simply 1λ that has made the two waves in phase.
At D2 above B1, a dark fringe is formed. Light coming from S2 travels a longer distance than the light coming from S1. The difference in the distance must be (3/2)λ that has made the two waves 180° out of phase.
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 (4/2)λ or simply 2λ that has made the two waves in phase.
The above repeating arguments can go on for as many bright and dark fringes as can be made. The same argument is also true for the fringes below Bo.
In practice, the two sources are very close, closer than one tenth of a millimeter. The distance between the two coherent sources is shown by d . The screen is also placed at least 1m from the slits. Light 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 called θ1. The angle that nearly both S2B2 and S1B2 make with the central line is name θ2 . See Figure below (top view). The fringes on the other side of the central line are not shown due to symmetry.
Young's Formula for the angle of the m-th Bright Fringe:
| It is easy to derive a formula
for the mth bright fringe. The formula relates
, d, and λ
, the wavelength of the light used. The formula is:
where m = 0, 1, 2, 3, ...
In the proof, note that the two angles that are labeled θm are equal because their sides are normal to each other.
|Let's magnify the circled area in the above figure
only for the nth bright fringe for which the distance difference
S2A = S2Bm - S1Bm = mλ.
In the right triangle S1S2A, sine of angle θm is:
sin θm = mλ /d as was expected.
Note that S1Bm and S2Bm 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 an 2.4° angle with respect to the centerline. The wavelength of the monochromatic light used is 480nm. Find the distance between the two sources.
Solution: sin θm = mλ /d ; d = mλ /sin θm ; d = 3(480x10-9) /sin(2.4°) = 34.4x10-6m = .034mm.
Young's Formula for the Angle of the m-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 m = 1, 2, 3, ... Note that there is no central dark fringe; therefore m =1 corresponds to the 1st dark fringe on each side.
Example 2: Find the angle for the 5th dark fringe in a double slit experiment with respect to the centerline knowing that the wavelength used is 550nm and the sources are 0.022mm apart.
Solution: sin θm = (m - 1/2)λ /d ; sin θ5 = (5 -1/2)(550x10-9m) /(0.022x10-3m) ; θ5 = 6.5°.
Alternate Method to measure angle θm:
If D, the distance from the slits to the screen or wall, is known, it is possible to measure ym, the distance from the central fringe to a desired mth fringe, and use a tangent function to calculate θm , the angle corresponding to that mth fringe.
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 7520mm from the screen. If the distance between the slits is 0.560mm, determine the wavelength and frequency of the light used.
Solution: tan (θm) = ym / D ; tan (θm) = 23.0 / 7520 ; θm = 0.175°.
sin θm = mλ /d ; sin 0.175° = 4λ /(0.560x10-3m) ; λ = 428x10-9m = 428nm.
c = f λ ; f = c /λ ; f = (3.00x108 m/s) / (428x10-9m) = 7.00x1014 Hz.
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 fringe?
Solution: sin θm = mλ /d ; sin θm = m (655x10-9m) / (5.00x10-6m) ; sin θm = 0.131m
Giving m different values, we may determine the number of possible fringes as in the following chart:
As can be seen, giving different values to m in
sin θm = 0.131m, results in 7 possible angles at which bright fringes occur. For m = 8, we don't get a fringe, because sine of an angle can not be more than 1.
An easier way to do this is to set sin θm = 1, this means setting 0.131m = 1 from which we get:
m = 1 / 0.131 = 7.64. Rounding to the lower integer, we get: m = 7.
|0||0||0 (Central Bright)|
|8||1.048 > 1||Not Possible|
Chapter 37 Test Yourself 1:
1) Two waves completely cancel each others effects if they have the same amplitude and are (a) 90° out of phase (b) 180° out of phase (c) 360° 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 at that point, 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 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) mλ, where m = 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) (m-1/2)λ, where m = 1, 2, 3, ....
11) The gap 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 θm , the angle corresponding to the m-th bright or dark fringe. (a) True (b) False
14) The angle θm corresponding to the m-th bright fringe is (a) sin θm = mλ (b) dsin θm = mλ (c) a & b. Of course, m = 0, 1, 2, 3, ... . click here.
15) The angle θm corresponding to the m-th dark fringe is (a) dsin θm = (m-1/2)λ (b) dsin θm = mλ (c) a & b. Of course m = 1, 2, 3, ... .
16) θm may also be calculated from the equation tanθm = ym /D2 where ym is the distance from the central bright fringe to the m-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 0.50° 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.
Problem: 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.
Important: In the following study, note that when a mechanical wave in a medium hits a harder medium, the reflected wave is 180° or λ /2 out of phase with respect to the incident wave. When a mechanical wave traveling in a medium hits a softer medium, the wave in the new medium remains in phase with incident wave. This is also true for light waves arriving at a new medium. If the new medium is optically harder (has a greater refraction index), the reflected wave will be completely out of phase with respect to the incident wave. When light waves enter an optically softer ( a lower refraction index) medium, they enter without a phase difference.
|A thin film of a liquid (gasoline or oil for example) floating on
another liquid (water for example) creates beautiful colorful patterns that
we all have noticed. The phenomenon is caused by the interference of
the immediately reflected rays from the top surface of the floating film and
reflected rays coming from the interface between water and gasoline.
See the figure shown.
a) The reflected ray 1 at A is λ /2 out of phase compared to IA.
b) The refracted ray that travels from A to B and then from B to C with a shorter wavelength in gasoline, of course, creates some phase difference as well.
Reflection at A causes λ /2 phase difference, but reflection at B that is at an optically softer medium occurs without a phase difference.
Since ngasoline > nair, gasoline acts as a hard surface for the incident ray IA, and the reflected ray 1 will be 180° out of phase with respect to the incident ray IA. The ray that refracts and enters gasoline will be in phase with the incident ray IA. When ray AB in gasoline hits the water surface, it reflects and travels the gasoline layer up and refracts back into air and forms ray 2 in the figure. This reflection at B from the water surface, does not cause a phase change with respect to the incident ray AB. The reason is that water is optically less dense than gasoline (nwater< ngasoline). All wavelengths in gasoline are shorter than those in air because gasoline is optically denser and slows down light. ( λgas. = λair / ngas.. Note that air and vacuum have very close optical properties.) This wavelength decrease in gasoline along with the thickness of the film puts the reflected ray from B out of phase with respect to the original incident ray IA.
In brief, we have reflected ray 1 that is 180° out of phase with respect to IA. The Reflected ray 2 is also somewhat out of phase with respect to IA. This means that in most cases reflected ray 1 and ray 2 are somewhat out of phase. The degree of being out of phase can vary from zero to 180° 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 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. It sometimes happens that for a certain thickness of the gasoline layer and angle of incidence a certain color is more pronounced or may be completely eliminated. That is when the reflected rays 1 and 2 are either completely in phase or completely out of phase.
Example 5: To an observer looking at an angle that is almost straight down, the color of a thin film of gasoline appears bright yellow. The wave length of yellow light in vacuum is 580.nm. 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 incident at small angles (close to their respective normal lines). The wavelength of yellow light in gasoline ( n = 1.40 ) is ( λgasoline = 580.nm / 1.40 = 413.5 nm ). The observer sees bright yellow light. It means that both reflected rays 1 and 2 are in phase and interfere constructively. This means that a distance difference equal to 2t along with a phase difference of 180° ( or λ /2 ) has caused the two reflected rays to be in phase again. We may write:
2t - λ /2 = mλ
where m = 0,1, 2, 3, ....
2t = λ /2 + mλ.
t = λ /4 + mλ /2.
|Case||m||t = λ /4 + mλ / 2.|
|1||0||t = λ /4 = 414nm /4 = 104nm.|
|2||1||t = 3λ /4 = 3(414nm) /4 = 310nm.|
|3||2||t = 5λ /4 = 5(414nm) /4 = 518nm.|
Note: If you write 2t + λ /2 = mλ, then you need to set m = 1, 2, 3, ... rather than m =0, 1, 2, ...
| An air wedge is the space between two sheets of glass
(or transparent media) that make a
small angle with each other. As the figure shows, the actual air wedge
is limited by the two solid lines. At the top line, light already
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 phase difference λ /2 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 constructive interference is:
2t + λ /2 = mλ, where m = 1, 2, 3, ..., or
2t + λ /2 = 1λ, 2λ, 3λ, ...
| The bending of light waves at sharp edges and openings is called
"diffraction." From a tiny point source like S as shown, we
expect to see a lit region on the wall only.
The following figures show the diffraction phenomenon. In Fig. 10, light waves from source S arrive at opening AB. Each air molecule at the opening starts to oscillate according to the wave it receives and becomes an independent source. It then sends waves out in all directions. The waves from these independent sources will have their constructive interferences in the R1 region with less constructive interferences in the semi-lit regions R2 and R3. R2 and R3 regions contain bright and dark fringes and are not uniform. This is just to bring your attention to the interference of a large number of independent sources.
In Fig. 11, the light passing a sharp edge at B has only one semi-lit region on the opposite wall. In both figures, it appears that sightline SB has bent its trajectory to form region R3.
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 film that some 5000 lines per inch is drawn on it by a special technique. When light passes through a diffraction grating, the grating becomes a place at which a large number of coherent sources are lined up. The space between every two lines on the grating becomes an independent source. The interference patterns of these sources can then be observed on the wall. In this case, there is a number of principal maxima (bright fringes) with many more weaker maxima in between. 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 mth principal maxima is similar to the one for Young's double-slit experiment. The N coherent sources and the relative intensity of the principal maxima are shown below:
| The formula that calculates the angles
corresponding to the principal maxima is
where m = 0, 1, 2, 3, ... correspond to the central, 1st, 2nd, 3rd, and ... principal maxima. In most cases, the central, 1st, and 2nd principal fringes are bright enough to be seen, the rest are normally very dim.
As the above figure shows, the central maxima is very bright. The intensities of the first, second, and third maxima on each side drop sharply.
Example 6: A diffraction grating is positioned 120.0cm from a wall and forms interference fringes as a 635nm laser passes through it. The very closely-spaced lines on the grating ( film) 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.
Solution: tan (θm) = ym / D ; tan (θm) = 31.5cm / 120.0cm ; θm = 14.71°.
sin θm = mλ /d ; sin (14.71°) = 2(635x10-9m) / d ; d = 5.00 x10-6m = 5.00x10-3mm / line.
To find the # of lines per mm, we need to reciprocate d. ; # of lines / mm = 1 / (5.00x10-3mm / line).
# of lines / mm = 200.
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. 8 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) in 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 and C2 is not exactly 180°, 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. 8 and repeat answering the questions. 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) 180° out of phase with IA (c) partially out of phase with IA.
16) The reflection at A is (a) in phase with IA. (b) 180° out of phase with IA. (c) partially out of phase with IA.
17) The refracted ray 1 is (a) in phase with IA (b) 180° 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) 180° 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) 180° 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) 180° out of phase with AB (c) partially out of phase with AB.
21) Rays CD and D2 are (a) in phase with AB (b) 180° 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) 180° out of phase with 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) 180° 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 dark and bright fringes as seen by an observer that is looking almost straight down onto the glass and 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 openings or by sharp edges. click here.
26) A diffraction grating is (a) a thin sheet of transparent material the has a large number of closely-spaced lines per inch (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 m-th principal maximum with respect to the central one that a diffraction grating forms is (a) sin θm = mλ (b) sin θm = λ /d (c) sin θm = mλ/d.
30) In the formula sin θm = mλ/d, or dsin θm = mλ, d is (a) the width of the grating (b) the space between every two adjacent lines (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 dsin θm = mλ 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 dsin θm = mλ for principal maxima of the grating's pattern, the wavelength of red light is (a) 404nm (b) 511nm (c) 688nm. click here.