Diffraction and Interference of Water Waves
Diffraction of Waves Diffraction the bending and spreading of a wave when it passes through an opening or around an obstacle Examples: sound waves travel through a door into the classroom or hall (due to long λ) water waves bend around a breakwall Can light bend too???
Diffraction of Water Waves Important to understand wave properties to identify wave phenomena Reasoning is since waves exhibit diffraction, so if a certain unexplained phenomenon exhibits diffraction, it must have wavelike properties
Diffraction of Water Waves What affects the amount of diffraction? wavelength (λ) longer λ = more diffraction frequency (f) lower f = more diffraction opening size (w) smaller opening (w)= more diffraction
Diffraction of Water Waves For waves to undergo noticeable diffraction, the wavelength must be comparable to or greater than the slit width where λ w As λ, diffraction As w, diffraction Ripple tank time
Examples Example 1: AM radio waves bend around hills much better than FM radio waves. What conclusion can you draw about the difference between AM and FM radio waves? AM = long wavelength, low frequency FM = short wavelength, higher frequency Example 2: Why can you hear a person standing behind a tree but you cannot see him/her? Sound waves are able to diffract around the tree due to their long wavelength but light has a very short wavelength due to high frequency and cannot diffract to make a person visible. BUT is light a wave???...
Interference of Waves Principle of Superposition when 2 waves in the same medium interact the resulting wave has an amplitude equal to the sum of the amplitudes of the 2 interfering waves Constructive Interference = larger amplitude (antinode) waves are in phase (same λ) Destructive Interference = smaller amplitude (node) waves are out of phase (λ differs) Refer to diagrams pg.462 *in both cases, the resultant amplitude of 2 waves A and B is the algebraic sum A+B
Interference of Waves Tap a dish of water in the centre Tap in 2 spots with constant f Note lines of constructive interference (antinodal lines) and lines of destructive interference (nodal lines) Refer to diagram pg. 463 Note: In the 2 source pattern although the nodal lines appear to be straight, their paths are actually hyperbolas
Interference of Waves Two Point Source Pattern: A pair of identical point sources that are in phase produce a symmetrical pattern of constructive interference areas and nodal lines. As the frequency of vibration of the sources and λ the nodal lines come closer together and the # of nodal lines
Interference of Waves Interference patterns can be used to analyze wave properties such as speed, wavelength, and frequency To perform suitable calculations, the sources must have the same f and v, thus same λ LET S ANALYZE A TWO-POINT SOURCE PATTERN
Mathematical Analysis of Two-Point Source Interference
Two-Point Source Interference Pattern If two identical sources, S1 and S2, are vibrating in phase with same f and λ Then there are an equal # of nodal lines on each side of the right bisector. For Point P1 connected to S1 and S2, the difference between the 2 pts is 1 2 λ
For any point P on the n th nodal line the difference in path length is: P n S1 PnS2 n 1 2 P n S 1 = path length; distance from source to pt along a nodal line This relationship measures the wavelength of interfering waves from a point on a specific nodal line. *This is valid for large wavelengths and close distances. When Point P is far away from the sources or λ is too small, a new approach is needed
When P n is very far away compared to d between S1 and S2, P, lines P n S 1 and P n S 2 ~ parallel Line AS 2 forms a right-angle triangle w/ both lines as triangle S 1 S 2 A Difference in path length in terms of θ n : sinθ n = AS 1 d P n S sinθ n = n 1 2 and AS 1 = P n S 1 P n S 2 where 1 PnS2 2 1 n Variables n- nth nodal line S1-source 1 S2-source 2 Pn-point on nth nodal line d-distance between 2 sources L-distance from centre of sources to nodal point λ d therefore: Refer to derivation Pgs. 465-467 Path length difference x n -distance from bisector to nodal point θ n -angle between d and AS2 NOTE: θ n is the angle to the nth nodal line
The previous eq n is good when it is relatively easy to measure angle θ n to a respective nodal line such as in a ripple tank. But what about when the wavelength and distance between sources is very small and the nodal lines are close together? How can info. be calculated for a two-point interference pattern without measuring θ n directly?
From the previous diagram, as P, lines P n S 1 and P n C ~ parallel which are both perpendicular to AS2. Since the right bisector CB is perpendicular to S1S2, from similar triangles, θ n = θ n From previous diagram sinθ n = x n and sinθ L n = n 1 2 Since θ n = θ n equating the two formulas results in the equation: λ d x n L = n 1 2 λ d This eq n is valid for small wavelengths and small distance d between sources. Note: You do not need θ n.
Summary: Two-Point Source Interference Equations for Sources Vibrating In-Phase λ d Note: sinθ 1 (max # of nodal lines occurs when θ=90 on 1 side of pattern, double for total # of nodal lines in pattern) 1. Source wavelength: sinθ n = n 1 2 2. Nodal line number: x n L = n 1 2 This eq n is good for small wavelengths and distances between sources. Formula #1 can also be used to calculate: i) the maximum number of nodal lines (sinθ n = 1 if θ n = 90 ) ii) the location of the antinodes λ d
Example #1: Two point sources 5.0 cm apart, in phase and connected to the same generator, create interfering waves with a frequency of 8.0 Hz in a ripple tank. One point on the first nodal line is 10.0 cm from one source and 11.0 cm from the other. Find: the wavelength of the waves speed of the waves Example #2: The distance from the right bisector to the second nodal line in a two-point interference pattern is 8.0 cm. The distance from the midpoint between the two sources to point P is 28 cm. What is the angle for the second nodal line?
EXTRA Example #1: Two point sources, in phase and connected to the same generator, create waves with a frequency of 2.5 Hz. One point on the third nodal line is 7.5 cm from one source and 15.0 cm from the other. Find: the wavelength speed of the wave Example #2: Two identical sources, in phase, create an interference pattern. One point on the 2nd nodal line is 3.80 cm from one source and 10.55 cm from another. How far apart should the two sources be placed so that they are situated at nodes with exactly four nodes between them.
Example #3: A two point source interference pattern is analyzed from a distance. A point on the third nodal line is 25cm from the centre of the two sources and 5.0cm from the median bisector. If the two sources are 2.5cm apart, find: the wavelength of the sources the angle between the bisector and nodal point