Interference of Waves Beats Double Slit Physics 1B03summer-Lecture 10

Physics 1B03summer-Lecture 10 Beats Two waves of different frequencies arriving together produce a fluctuation in power or amplitude. Since the frequencies are different, the two vibrations drift in and out of phase with each other, causing the total amplitude to vary with time. y time 1 beat Physics 1B03summer-Lecture 10

Physics 1B03summer-Lecture 10 time in phase 180o out of phase in phase t Physics 1B03summer-Lecture 10

Physics 1B03summer-Lecture 10 The math: Same amplitudes, different frequencies: Trigonometry: cos a + cos b = 2 cos [(a-b)/2] cos [(a+b)/2] Result: slowly-varying amplitude SHM at average frequency Physics 1B03summer-Lecture 10

Physics 1B03summer-Lecture 10 Note: 2 beats per cycle of # beats/second = The beat frequency (number of beats per second) is equal to the difference between the frequencies: Physics 1B03summer-Lecture 10

Physics 1B03summer-Lecture 10 Quiz Two guitar strings originally vibrate at the same 400-Hz frequency. If you hear a beat of 5Hz, what is/are the other possible frequencie(s) ? 10 Hz 395 Hz 405 Hz 395 Hz and 405 Hz Physics 1B03summer-Lecture 10

Physics 1B03summer-Lecture 10 Interference 2 waves, of the same frequency; arrive out of phase. Eg: y1=Asin wt y2=Asin (wt+f) Then yR= y1 + y2 = AR sin(wt+fR), and the resultant amplitude is AR=2Acos(½f). Identical waves which travel different distances will arrive out of phase and will interfere, so that the resultant amplitude varies with location. Physics 1B03summer-Lecture 10

Physics 1B03summer-Lecture 10 Phase difference : Define Then, at detector: (pick starting time so initial phase is zero here) Physics 1B03summer-Lecture 10

Physics 1B03summer-Lecture 10 Example: Two sources, in phase; waves arrive by different paths: S1 P r1 detector r2 At detector P: S2 Physics 1B03summer-Lecture 10

Physics 1B03summer-Lecture 10 detector x 2 speakers, in phase; f = 170 Hz (so l = 2.0 m; the speed of sound is about 340 m/s) As you move along the x axis, where is the sound: a minimum (compared to nearby points)? a maximum (compared to nearby points)? Physics 1B03summer-Lecture 10

Physics 1B03summer-Lecture 10 Solution: Physics 1B03summer-Lecture 10

Physics 1B03summer-Lecture 10 10 min rest Physics 1B03summer-Lecture 10

Physics 1B03summer-Lecture 10 Interference of Light Light is an electromagnetic (EM) wave. Wave properties: Diffraction – bends around corners, spreads out from narrow slits Interference – waves from two or more coherent sources interfere Physics 1B03summer-Lecture 10

Electromagnetic Waves B (magnetic field) Eo v Usually we keep track of the electric field E : Electric field amplitude Electromagnetic waves are transverse waves Physics 1B03summer-Lecture 10

Physics 1B03summer-Lecture 10 Infrared Red 780 nm Yellow 600 nm Green 550 nm Blue 450 nm Violet 380 nm Ultraviolet l Visible-Light Spectrum Physics 1B03summer-Lecture 10

The Electromagnetic Spectrum l (m) f (Hz) 300 106 3 108 3 x 10-3 1011 3 x 10-6 1014 7 x 10-7 5x1014 4 x 10-7 3 x 10-9 1017 3 x 10-12 1020 Radio TV Microwave Infrared Visible Ultraviolet X rays g rays Physics 1B03summer-Lecture 10

Our galaxy (Milky Way) at viewed at different wavelengths Physics 1B03summer-Lecture 10

Physics 1B03summer-Lecture 10 Radio lobes (jets) from a supermassive black hole at the center of the galaxy NGC 4261 Physics 1B03summer-Lecture 10

Physics 1B03summer-Lecture 10 Double Slit (Thomas Young, 1801) m=2 m=1 m=0 (center) m=-1 m=-2 θ incident light double slit separation d screen Result: Many bright “fringes” on screen, with dark lines in between. Physics 1B03summer-Lecture 10

Physics 1B03summer-Lecture 10 The slits act as two sources in phase. Due to diffraction, the light spreads out after it passes through each slit. When the two waves arrive at some point P on the screen, they can be in or out of phase, depending on the difference in the length of the paths. The path difference varies from place to place on the screen. P r1 To determine the locations of the bright fringes (interference maxima), we need to find the points for which the path difference Dr is equal to an integer number of wavelengths. For dark fringes (minima), the path difference is integer multiples of half of a wavelength. r2 q d Δr = r1-r2 Physics 1B03summer-Lecture 10

Physics 1B03summer-Lecture 10 For light, the slits will usually be very close together compared to the distance to the screen. So we will place the screen “at infinity” to simplify the calculation. move P to infinity r1 P r >> d, r1 & r2 nearly parallel r2 d q θ Δr = r1-r2 d θ Δr Δr = d sin θ Physics 1B03summer-Lecture 10

Physics 1B03summer-Lecture 10 Interference: 2 coherent waves, out of phase due to a path difference Dr: Constructive Interference (maximum intensity) for f = 0, ±2π, ±4π, ±6π, ……… -> Δr =0, ±l, ±2 l, ±3 l, ……… Destructive Interference (minimum intensity) for f = ±π, ±3π, ±5π, ……… -> Δr =±λ/2, ±3λ/2, ±5λ/2, ……… Physics 1B03summer-Lecture 10

Physics 1B03summer-Lecture 10 Constructive Interference: (bright) Δr = mλ or d sin θ = mλ, m = 0, ±1, ±2, … But, if the slit-screen distance (L) is large, then sinθ~θ and so sinθ=θ=y/L (in radians): L y θ d So we have: Physics 1B03summer-Lecture 10

Physics 1B03summer-Lecture 10 Destructive Interference: (no light) Δr = (m + ½)λ or d sin θ = (m + ½) λ, m = 0, ±1, ±2, … So, we have: Physics 1B03summer-Lecture 10

Physics 1B03summer-Lecture 10 Quiz Two slits are illuminated with red light to produce an interference patter on a distant screen. If the red light is replaces with blue light, how does the pattern change? The bright spots move closer together The bright spots move farther apart The pattern does not change The patter doesn’t chance, but the width of the spots changes Physics 1B03summer-Lecture 10

2 slits, 0.20 mm apart; red light (l = 667 nm) Example 2 slits, 0.20 mm apart; red light (l = 667 nm) y 3 m screen Where are a) the bright fringes? b) the dark lines? (give values of y) Physics 1B03summer-Lecture 10

Physics 1B03summer-Lecture 10 Solution: Physics 1B03summer-Lecture 10

Physics 1B03summer-Lecture 10 Example A double slit interference patter is observed on a screen 1.0m behind two slits spaced 0.3mm apart. Ten bright fringes span a distance of 1.65 cm. What is the wavelength of light used ? Physics 1B03summer-Lecture 10

Physics 1B03summer-Lecture 10 Quiz Which of the following would cause the separation between the fringes to decrease? Increasing the wavelength Decreasing the wavelength Moving the slits closer together Moving the slits farther apart None of the above Physics 1B03summer-Lecture 10

Physics 1B03summer-Lecture 10 10 min rest Physics 1B03summer-Lecture 10

Physics 1B03summer-Lecture 10 Refractive Index The speed of light depends on the material. We define the refractive index “n” as n = (speed of light in vacuum)/(speed of light in a material) material refractive index speed of light vacuum 1 c  300,000 km/s air 1.0003 glass about 1.5 200,000 km/s water 1.333 225,000 km/s diamond 2.4 125,000 km/s Physics 1B03summer-Lecture 10

Physics 1B03summer-Lecture 10 Question: A beam of yellow light (wavelength 600 nm), travelling in air, passes into a pool of water. By what factor do the following quantities change as the beam goes from air into water? speed frequency wavelength Physics 1B03summer-Lecture 10

Reflection and Phase Change Light waves may have a 180° phase change when they reflect from a boundary: “optically dense” medium (larger refractive index) no phase change at this reflection 180° phase change when reflecting from a denser medium Just remember this : low to high, phase shift of pi ! Physics 1B03summer-Lecture 10

Physics 1B03summer-Lecture 10 Example: Thin film What is the minimum thickness of a soap film (n  1.33) needed to produce constructive interference for light with a 500nm wavelength ? (air : n  1.00). What about destructive interference ? Physics 1B03summer-Lecture 10

Example: Antireflection coatings To reduce reflections from glass lenses (n  1.5), the glass surfaces are coated with a thin layer of magnesium fluoride (n  1.38). What is the correct thickness of the coating for green light (550 nm vacuum wavelength)? MgF2 air glass Physics 1B03summer-Lecture 10

Physics 1B03summer-Lecture 10 Example A beam of 580 nm light passes through two closely spaced glass plates (nglass=1.6), as shown in the figure below. For what minimum nonzero value of the plate separation d is the transmitted light dark? Physics 1B03summer-Lecture 10

Physics 1B03summer-Lecture 10 Quiz Why do we see many colours on a soap bubble? A) because white light is made up of different wavelengths B) because the bubble has different thickness C) both A and B D) because the bubble is round and light reflects from the other side Physics 1B03summer-Lecture 10