Double Slit Experiment and diffrac5on gra5ngs
Diffraction
Diffraction is normally taken to refer to various phenomena which occur when a wave encounters an obstacle. It is described as the apparent bending of waves around small obstacles and the spreading out of waves past small openings
Superposition, or Interference Another characteristic of a WAVE is the ability to undergo INTERFERENCE. There are TWO types. We call these waves IN PHASE. We call these waves OUT OF PHASE.
Interference applet
Double-Slit Experiment to illustrate wave nature of light
Thomas Young s Double Slit Interference Experiment Gave credibility to the theory that light is a wave Showed an interference pattern Measured the wavelength of the light
http://galileo.phys.virginia.edu/classes/usem/sciimg/home_files/introduction_files/doubleslit.jpg
Young s Double Slit Experiment
Young s Double Slit Interference Pattern
Interference Fringes on a Screen applet
Conditions Chapter 15 for Interference of Light Waves
Diffraction The Central Maximum Suppose you had TWO sources each being allowed to emit a wave through a small opening or slit. The distance between the slits denoted by, d. The distance from the slit spacing to the screen is denoted by the letter, L. If two waves go through the slit and then proceed straight ahead to the screen, they both cover the SAME DISTANCE and thus will have constructive interference. Their amplitudes will build and leave a very bright intense spot on the screen. We call this the CENTRAL MAXIMUM.
Diffraction The Central Maximum Figure 1 Figure 2 Here is the pattern you will see. Notice in figure 2 that there are several bright spots and dark areas in between. The spot in the middle is the BRIGHTEST and thus the CENTRAL MAXIMUM. We call these spots FRINGES. Notice we have additional bright spots, yet the intensity is a bit less. We denote these additional bright spots as ORDERS. So the first bright spot on either side of the central maximum is called the FIRST ORDER BRIGHT FRINGE. Figure 1 represents the intensity of the orders as we move farther from the bright central maximum.
Diffraction Orders, m We use the letter, m, to represent the ORDER of the fringe from the bright central. Second Order Bright Fringe m = 2 First Order Bright Fringe m = 1 Central Maximum m = 0 First Order Bright Fringe m = 1 Second Order Bright Fringe m = 2 It is important to understand that we see these bright fringes as a result of CONSTRUCTIVE INTERFERENCE. So bright fringes are ordered m = 0,1,2,3.
Diffraction Bright Fringes The reason you see additional bright fringes is because the waves CONSTRUCTIVELY build. There is a difference, however, in the intensity as you saw in the previous slide. As you can see in the picture, the BLUE WAVE has to travel farther than the RED WAVE to reach the screen at the position shown. For the BLUE WAVE and the RED WAVE to build constructively they MUST be IN PHASE. Here is the question: HOW MUCH FARTHER DID THE BLUE WAVE HAVE TO TRAVEL SO THAT THEY BOTH HIT THE SCREEN IN PHASE?
Diffraction Path difference Notice that these 2 waves are IN PHASE. When they hit they screen they both hit at the same relative position, at the bottom of a crest. λ How much farther did the red wave have to travel?
Diffraction Path difference Notice that these 2 waves are IN PHASE. When they hit they screen they both hit at the same relative position, at the bottom of a crest. λ How much farther did the red wave have to travel? ONE WAVELENGTH
Diffraction Path difference Notice that these 2 waves are IN PHASE. When they hit they screen they both hit at the same relative position, at the bottom of a crest. λ How much farther did the red wave have to travel? ONE WAVELENGTH We call this extra distance the PATH DIFFERENCE. The path difference and the ORDER of a fringe help to form a pattern.
Diffraction Path Difference 2λ 1λ 1λ 2λ The bright fringes you see on either side of the central maximum are multiple wavelengths from the bright central. And it just so happens that the multiple is the ORDER. Therefore, the PATH DIFFERENCE is equal to the ORDER times the WAVELENGTH P.D. = mλ (Constructive)
Diffraction Dark Fringes We see a definite DECREASE in intensity between the bright fringes. In the pattern we visibly notice the DARK REGION. These are areas where DESTRUCTIVE INTERFERENCE has occurred. We call these areas DARK FRINGES or MINIMUMS (minima).
Diffraction Dark Fringes First Order Dark Fringe m = 1 ZERO Order Dark Fringe m = 0 ZERO Order Dark Fringe m = 0 First Order Dark Fringe m = 1 It is important to understand that we see these dark fringes as a result of DESTRUCTIVE INTERFERENCE. So dark fringes are ordered m = 0,1,2,3.
Diffraction Dark Fringes On either side of the bright central maximum we see areas that are dark, or minimum intensity. Once again, we notice that the BLUE WAVE had to travel farther than the RED WAVE to reach the screen. At this point, however, they are said to destructively build or that they are OUT OF PHASE. Here is the question: HOW MUCH FARTHER DID THE BLUE WAVE HAVE TO TRAVEL SO THAT THEY BOTH HIT THE SCREEN OUT OF PHASE?
Diffraction Path difference Notice that these 2 waves are OUT OF PHASE. When they hit they screen they both hit at the different relative positions, one at the bottom of a crest and the other coming out of a trough. Thus their amplitudes SUBTRACT. 0.5 λ How much farther did the red wave have to travel?
Diffraction Path difference Notice that these 2 waves are OUT OF PHASE. When they hit they screen they both hit at the different relative positions, one at the bottom of a crest and the other coming out of a trough. Thus their amplitudes SUBTRACT. 0.5 λ How much farther did the red wave have to travel? ONE HALF OF A WAVELENGTH
Diffraction Path difference Notice that these 2 waves are OUT OF PHASE. When they hit they screen they both hit at the different relative positions, one at the bottom of a crest and the other coming out of a trough. Thus their amplitudes SUBTRACT. 0.5 λ How much farther did the red wave have to travel? ONE HALF OF A WAVELENGTH The call this extra distance the PATH DIFFERENCE. The path difference and the ORDER of a fringe help to form another pattern.
Diffraction Path Difference 1.5λ 1.5λ 0.5λ 0.5λ The dark fringes you see on either side of the central maximum are multiple wavelengths from the bright central. And it just so happens that the multiple is the ORDER. Therefore, the PATH DIFFERENCE is equal to the ORDER plus a HALF, times A WAVELENGTH P.D. = (m+1/2)λ (Destructive)
Path Difference - Summary For CONSTRUCTIVE INTERFERENCE or MAXIMA use: P.D. = mλ Where m = 0, 1, 2, 3 For DESTRUCTIVE INTERFERENCE or MINIMA use: P.D. = (m + 1/2) λ Where m, is the ORDER. Where m = 0, 1, 2, 3
d Young s Experiment In 1801, Thomas Young successfully showed that light does produce an interference pattern. This famous experiment PROVES that light has WAVE PROPERTIES. Suppose we have 2 slits separated by a distance, d and a distance L from a screen. Let point P be a bright fringe. θ P y central max We see in the figure that we can make a right triangle using, L, and y, which is the distance a fringe is from the bright central. We will use an angle, θ, from the point in the middle of the two slits. L We can find this angle using tangent!
Similar Triangles (or more often, y )
Diffraction Another way to look at path difference P This right triangle is SIMILAR to the one made by y & L. θ d θ central max d P.D. P.D. = dsinθ P.D. Notice the blue wave travels farther. The difference in distance is the path difference.
(Note that for a large distance, L, from the screen, ray 1 and ray 2 are approximately parallel).
Similar Triangles lead to a path difference y θ These angles Are EQUAL. L d dsinθ θ
sinθ = opp/hyp = path length difference / d So path length difference = d sinθ
Diffraction Putting it all together Path difference is equal to the following: mλ for constructive interference (m+1/2)λ for destructive interference dsinθ for both Therefore, we can say: Constructive Interference dsinθ = mλ Will be used to find the angle!
Similar Triangles (or more often, y ) Notice: tan θ = y / L
Example #1 A viewing screen is separated from a double slit source by 1.2 m. The distance between the two slits is 0.030 mm. The second -order bright fringe ( m=2) is 4.5 cm from the central maximum. Determine the wavelength of light.
Example #1 A viewing screen is separated from a double slit source by 1.2 m. The distance between the two slits is 0.030 mm. The second -order bright fringe ( m=2) is 4.5 cm from the central maximum. Determine the wavelength of light. 2.15 degrees dsinθ = mλ (3x10 5 )sin(2.15) = 2λ λ = 5.62x10-7 m
Small Angle Approximation When L is large, the hypotenuse is very close in value to L If θ is small (less than a few deg), we can say that sinθ tanθ
L d sinθ = mλ m = 0,1,2,3... Maximum d sinθ d tanθ = dy L = mλ also, tanθ = y m L so for maxima, or y m = mλl d Maxima y m = Ltanθ Lsinθ Minima m y m (+/-) m y m (+/-) 0 1 2 3 0 Lλ/d 2Lλ/d 3Lλ/d 0 1 2 3 Lλ/2d 3Lλ/2d 5Lλ/2d 7Lλ/2d (or P.D.) & for minima, = d sinθ dsinθ = (m + 1 )λ m = 0,1,2,3... Minimum 2 y m = (m +1/2)λL d
EXAMPLE #2 Suppose that Young s experiment is performed with blue-green light of 500 nm. The slits are 1.2 mm apart, and the viewing screen is 5.4 m from the slits. How far apart are the bright fringes?
EXAMPLE #2 Suppose that Young s experiment is performed with blue-green light of 500 nm. The slits are 1.2 mm apart, and the viewing screen is 5.4 m from the slits. How far apart are the bright fringes? From the table on the previous slide we see that the separation between bright fringes is Δy =
EXAMPLE #2 Suppose that Young s experiment is performed with blue-green light of 500 nm. The slits are 1.2 mm apart, and the viewing screen is 5.4 m from the slits. How far apart are the bright fringes? From the table on the previous slide we see that the separation between bright fringes is Δy = Lλ /d
EXAMPLE #2 Suppose that Young s experiment is performed with blue-green light of 500 nm. The slits are 1.2 mm apart, and the viewing screen is 5.4 m from the slits. How far apart are the bright fringes? From the table on the previous slide we see that the separation between bright fringes is Δy = Lλ /d Lλ /d = (5.4m)(500 10 9 m) /0.0012m
EXAMPLE #2 Suppose that Young s experiment is performed with blue-green light of 500 nm. The slits are 1.2 mm apart, and the viewing screen is 5.4 m from the slits. How far apart are the bright fringes? From the table on the previous slide we see that the separation between bright fringes is Δy = Lλ /d Lλ /d = (5.4m)(500 10 9 m) /0.0012m = 0.00225m = 2.25mm
A diffraction grating that consists of a large number of parallel slits overcomes both of these difficulties. A diffraction grating uses interference to disperse light. It is often an important component in optical instrumentation for wavelength determinations.
For a diffraction grating, the intensity falls away from these maxima much more rapidly than that for a double slit. Because there are so many slits to act as sources, any angle other than those for maxima will be dark or nearly dark.
Diffraction gratings A diffraction grating is a precise array of tiny engraved lines, each of which allows light through. The spectrum produced is a mixture of many different wavelengths of light.
How a Diffraction Grating Works When you look at a diffracted light you see: the light straight ahead as if the grating were transparent. a "central bright spot". the interference of all other light waves from many different grooves produces a scattered pattern called a spectrum.
Note: We cannot use the small angle approximation for diffraction gratings. For gratings, since d is very small, to observe some of the higher order fringes, we usually have to place the screen very close to the grating; therefore, the angle θ is no longer small.
The Diffraction Grating Diffraction grating is an arrangement consisting of a large number of parallel, closely spaced slits. Gratings with as many as 40 000 slits per centimeter can be made, depending on the production method. In one method a diamond-tipped cutting tool is used to inscribe closely spaced parallel lines on a glass plate, the spaces between the lines serving as the slits.
Double Slit & Grating The bright fringes produced by a diffraction grating are much narrower than those produced by a double slit. Note the three small secondary bright fringes between the principal bright fringes of the grating. For a large number of slits, these secondary fringes become very small.
Grating Formula wavelength of light (nm) distance between grating lines (m) mλ = d sinθ constructive d sinθ = (m + 1 2 )λ destructive Note: sometimes the given data for slit separation is given as lines per _ meters instead of d Set 1/d = number of lines per m (NOTE: a line is the solid material separating each slit).
Double Slit & Grating The bright fringes produced by a diffraction grating are much narrower than those produced by a double slit. Note the three small secondary bright fringes between the principal bright fringes of the grating. For a large number of slits, these secondary fringes become very small.
Example #3 A light with wavelength, 450 nm, falls on a diffraction grating (multiple slits). On a screen 1.80 m away the distance between dark fringes on either side of the bright central is 4.20 mm. a) What is the separation between a set of slits? b) How many lines per meter are on the grating?
Example #3 A light with wavelength, 450 nm, falls on a diffraction grating (multiple slits). On a screen 1.80 m away the distance between dark fringes on either side of the bright central is 4.20 mm. a) What is the separation between a set of slits? b) How many lines per meter are on the grating? NOTE: a line is the solid material separating each slit.
Example #3 A light with wavelength, 450 nm, falls on a diffraction grating (multiple slits). On a screen 1.80 m away, the distance between dark fringes on either side of the bright central is 4.20 mm. a) What is the separation between a set of slits? b) How many lines per meter are on the grating? NOTE: a line is the solid material separating each slit. λ = 450x10 9 m L =1.80m y = 4.2mm /2 = 0.0021m θ =? d =? 0.067 degrees 5197.2 lines/m 0.0001924 m