Interference ApplicationsPAL #23 InterferenceSlide 3Slide 4Slide 5Slide 6Slide 7OrdersIntensity of Interference PatternsIntensityIntensity VariationThin Film InterferenceReflection Phase ShiftsReflection and Thin FilmsPath Length and Thin FilmsReflection and InterferenceInterference DependenciesColor of FilmNext TimeInterference: SummaryReflectionPath Length DifferenceDifferent Index of RefractionInterference ApplicationsPhysics 202Professor Lee CarknerLecture 25PAL #23 InterferenceLight with  = 400 nm passing through n=1.6 and n=1.5 materialN = (L/)(n)L = N /n = (5.75)(400)/(0.1) = 23000 nmCompare to L = 2.6X10-5 mN = (2.6X10-5)(0.1)/(400X10-9) = 6.56.5  is total destructive interference and so the above situation is brighter (5.75 )What directions will the beam be bent towards as it enters A, B and C?a) Up, up, upb) Down, down, downc) Up, down, upd) Up, up, downe) Down, up, down A B Cn=1n=1.4n=1.3n=1.5Rank the 3 materials by the speed of light in them, greatest first.a) A, B, Cb) B, C, Ac) C, A, Bd) A, C, Be) Speed is the same in all A B Cn=1n=1.4n=1.3n=1.5What happens to the distance between the fringes if the distance between the slits increases?a) Increasesb) Decreasesc) Stays the sameWhat happens to the distance between the fringes if the light is switched from red to green?a) Increasesb) Decreasesc) Stays the sameWhat happens to the distance between the fringes if the entire apparatus in submerged in a clear liquid?a) Increasesb) Decreasesc) Stays the sameOrders At the center is the 0th order maxima, flanked by the 0th order minima, next is the 1st order maxima etc.The orders are symmetric e.g. the 5th order maxima is located both to the left and the right of the center at the same distanceThe intensity varies sinusoidally between minima and maximaIntensity of Interference PatternsHow bright are the fringes?  The phase difference is related to the path length difference and the wavelength and is given by: = (2d sin ) / Where d is the distance between the slits, and  is the angle to the point in question is in radiansIntensityThe intensity can be found from the electric field vector E:I  E2I = 4 I0 cos2 (½ ) For any given point on the screen we can find the intensity if we know ,d, and I0 The average intensity is 2I0 with a maximum and minimum of 4I0 and 0Intensity VariationThin Film Interference Camera lenses often look bluish Light that is reflected from both the front and the back of the film has a path length difference and thus may also have a phase difference and show interferenceReflection Phase Shifts The phase shift depends on the relative indices of refractionIf light is incident on a material with lower n, the phase shift is 0 wavelengthExample: If light is incident on a material with higher n, the phase shift is 0.5 wavelengthExample: The total phase shift is the sum of reflection and path length shiftsReflection and Thin Films Since nfilm > nair and nglass > nfilm Example: optical antireflection coatings Since nfilm > nair and nair < nfilmHave to add 0.5 wavelength shift to effects of path length differenceExample: soap bubblePath Length and Thin FilmsFor light incident on a thin film, the light is reflected once off of the top and once off of the bottom If the light is incident nearly straight on (perpendicular to the surface) the path length difference is 2 times the thickness or 2LDon’t forget to include reflection shiftsReflection and InterferenceWhat kind of interference will we get for a particular thickness? The wavelength of light in the film is equal to:2 = /n2For an anti-reflective coating (no net reflection shift), the two reflected rays are in phase and they will produce destructive interference if 2L is equal to 1/2 a wavelength2L = (m + ½) (/n2) -- dark film 2L = m (/n2) -- bright filmInterference DependenciesFor a film in air (soap bubble) the equations are reversed Soap film can appear bright or dark depending on the thickness Since the interference depends also on  soap films of a particular thickness can produce strong constructive interference at a particular This is why films show colorsColor of FilmWhat color does a soap film (n=1.33) appear to be if it is 500 nm thick?We need to find the wavelength of the maxima:2L = (m + ½) (/n) = [(2) (500nm) (1.33)] / (m + ½) = 2660 nm, 887 nm, 532 nm, 380 nm …Only 532 nm is in the visible region and is greenNext TimeRead: 36.1-36.6Homework: Ch 35, P: 40, 53, Ch 36, P: 2, 17Interference: SummaryInterference occurs when light beams that are out of phase combineThe interference can be constructive or destructive, producing bright or dark regionsThe type of interference can depend on the wavelength, the path length difference, or the index of refractionWhat types of interference are there?ReflectionDepends on: nExample: thin filmsEquations: •n1 > n2 -- phase shift = 0•antireflective coating•n1 < n2 -- phase shift = 0.5•soap bubblePath Length DifferenceDepends on: L and Example: double slit interferenceEquations: d sin  = m -- maximad sin  = (m + ½) -- minimaDifferent Index of RefractionDepends on: L, , nExample: combine beams from two mediaEquations: N2 - N1 = (L/)(n2