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Interference Applications Physics 202 Professor Lee Carkner Lecture 23
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PAL #22 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 )
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Interference Patterns For a double slit experiment of slit separation d, the position of the maxima and minima are: d sin = (m+½) For a maxima of order m, the linear distance from the central maxima on the screen: where m is the order number and D is the distance to the screen
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Intensity 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 ) / is in radians
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Intensity I E 2 I = 4 I 0 cos 2 (½ ) Where I 0 is the intensity of the direct light from one slit and is the phase difference in radians The average intensity is 2I 0 with a maximum and minimum of 4I 0 and 0
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Thin Film Interference Consider light that reflects off of the front and back of a thin film and recombines: There will be a phase shift due to the path length difference Combine all together to find total interference
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Reflection Phase Shifts 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 shifts
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Reflection and Thin Films Since n film > n air and n glass > n film Interference is due only to path length difference Example: optical antireflection coatings Since n film > n air and n air < n film Have to add 0.5 wavelength shift to effects of path length difference Example: soap bubble
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Anti-Reflective Coating The wavelength of light in the film is equal to: Since there is no net reflection shift, the two reflected rays are in phase and they will produce destructive interference if 2L is equal to ½ wavelength 2L = (m + ½) ( /n 2 ) -- dark film 2L = m ( /n 2 ) -- bright film
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Soap Film Since the net reflection shift is ½ wavelength, the two reflected rays will produce constructive interference if 2L is equal to ½ wavelength 2L = (m + ½) ( /n 2 ) -- bright film 2L = m ( /n 2 ) -- dark film If a certain wavelength produces bright film, the film will be that color
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Next Time Read: 36.1-36.6
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What directions will the beam be bent towards as it enters A, B and C? a)Up, up, up b)Down, down, down c)Up, down, up d)Up, up, down e) Down, up, down ABC n=1 n=1.4n=1.3n=1.5
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Rank the 3 materials by the speed of light in them, greatest first. a)A, B, C b)B, C, A c)C, A, B d)A, C, B e)Speed is the same in all ABC n=1 n=1.4n=1.3n=1.5
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What happens to the distance between the fringes if the distance between the slits increases? a)Increases b)Decreases c)Stays the same
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What happens to the distance between the fringes if the light is switched from red to green? a)Increases b)Decreases c)Stays the same
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What happens to the distance between the fringes if the entire apparatus in submerged in a clear liquid? a)Increases b)Decreases c)Stays the same
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Interference: 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?
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Reflection Depends on: n Example: thin films Equations: n 1 > n 2 -- phase shift = 0 antireflective coating n 1 < n 2 -- phase shift = 0.5 soap bubble
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Path Length Difference Depends on: L and Example: double slit interference Equations: d sin = m -- maxima d sin = (m + ½) -- minima
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Different Index of Refraction Depends on: L,, n Example: combine beams from two media Equations: N 2 - N 1 = (L/ )(n 2 -n 1 )
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