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# OSU PHYSICS 1251 - Example Problem: Interference Part 2

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Example Problem: Interference Part 2Physics 1251TA: Brian Clark11/23/20151. Two speakers are coherent sources of sound that emit sound waves in phase. PointP is a distance d1= 2.45 m from one speaker and a distance d2= 1.78 m from the otherspeaker. The speed of sound is 343 m/s. Find three different frequencies that willexperience destructive interference at P .For destructive destructive interference at P , we want the path length difference between eachof the speakers and P to be equal to some odd half-integer multiple of wavelength λ.∆d = d2− d1= (m + 1/2)λ; m = 0, 1, 2, ...Additionally, we know that v = fλ, so∆d =(m + 1/2)vfSolving for f, we find:(m = 0) f0=v2∆d= 256 Hz (m = 1) f1=3v2∆d= 768 Hz (m = 2) f2=5v2∆d= 1280 Hz2. Light of wavelength λ = 600 nm passes through a pair of slits; each slit is 50 µm wideand the slits are separated by 170 µm.(a) How many bright fringes occur on a screen 1.3 m wide that is 1.5 m past the slits?We want to count the total number of bright fringes in the double slit pattern out to theangle θ, where the end of the screen is. The screen is 1.5 m away from the slits and the edgeof the screen is located at a distance of 1.3/2 = 0.65 m from its center, soθ = arctan0.651.5= 23.4◦Next we can use the equation representing double slit interference maxima to calculate theorder m of the last bright fringe on the screen:dsinθ = mλ =⇒ m =dsinθλ=⇒ m = 112.7This means there are 112 fringes to the right, and to the left, of the central bright maximumat m = 0. So in total, we have:N = 112 + 112 + 1 = 225 fringes(b) How many bright fringes occur in the whole pattern?The last bright fringe in the entire pattern occurs when the angle of interference is θ = 90◦.Thus, we set sinθ = 1 in the double slit interference maximum equation and find the order ofthe last interference maximum:d · 1 = mλ =⇒ m =170µm600µm= 283.3Meaning there are 283 full fringes on each side of the diffraction pattern, and then the one inthe middle. So the total in the pattern isN = 283 + 283 + 1 = 567 fringes3. Two colors of light pass through a double slit at the same time. The second interfer-ence minimum of color A is at the same angle as the third maximum of color B. Whatis the ratio of λA/λB?The equations describing interference maxima and minima for a double slit are:d sin θ = mλ (maxima)d sin θ = (m − 1/2)λ (minima)Since the second minimum of color A and the third maxima of color B are located at the sameangle, we can write:d sin θ =32λA& d sin θ = 3λB32λA= 3λB=⇒λAλB= 24. Two colors of light pass through a single slit at the same time. The second diffractionminimum of color A is at the same angle as the third maximum of color B. What isthe ratio of λA/λB?The equations describing interference maxima and minima for a single slit are:d sin θ = mλ (minima)d sin θ = (m + 1/2)λ (maxima)Again, the second minimum of color A is at the same angle as the third maximum of color B:d sin θ =72λB& d sin θ = 2λA=⇒72λA= 2λB=⇒λAλB=74= 1.755. A 0.500 micrometers thick film of n = 1.25 soap is suspended in mid-air.(a) For what wavelengths of visible light is the reflected light a maximum? For whatwavelengths of visible light is the reflected light a minimum?We are going to employ the equations for constructive and destructive thin film interferencein the case where n3(air) < n2(soap) > n1(air). Then we have:2t = (m + 1/2)λn(maxima) & 2t = mλn(minima)For maxima, the only m with a λ in the visible spectrum (400 − 700 nm) is m = 2 =⇒λ2= 500 nm . For minima, m = 2 and m = 3 both work with m = 2 =⇒ λ2= 625 nm andm = 3 =⇒ λ3= 416 nm .(b) If instead of being suspended in mid-air, the soap film was applied to a piece ofglass (with, say, nglass= 1.5), what would those wavelengths be?If the film as applied to glass, then now, n3(glass) > n2(soap) > n1(air), so the condi-tions for destructive and constructive interference change, and our answers above just swaproles. Maxima are now found at 625 nm and 416 nm, and minima at 500

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