Download presentation
Presentation is loading. Please wait.
1
Interference Applications Physics 202 Professor Lee Carkner Lecture 25
2
PAL #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) = Compare to L = 2.6X10 -5 m N = (2.6X10 -5 )(0.1)/(400X10 -9 ) = 6.5
3
Orders At the center is the 0th order maxima, flanked by the 0th order minima, next is the 1st order maxima etc. The intensity varies sinusoidally between minima and maxima
4
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
5
Intensity The intensity can be found from the electric field vector E: I = 4 I 0 cos 2 (½ ) For any given point on the screen we can find the intensity if we know ,d, and I 0 The average intensity is 2I 0 with a maximum and minimum of 4I 0 and 0
6
Intensity Variation
7
Thin 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 interference
8
Thin Film
9
Reflection Phase Shifts In addition to the path length shift there can also be a phase shift due to reflection 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
10
Reflection Phase Change
11
Reflection and Thin Films If the thin film covers glass, both reflection phase shifts will be 0.5 Interference is due only to path length difference Example: If the thin film is in air, the first shift is 0.5 and the second is zero Have to add 0.5 wavelength shift to effects of path length difference Example:
12
Path Length and Thin Films What is the path length difference between the two reflected rays? Don’t forget to include reflection shifts
13
Anti-reflective Coating
14
Reflection and Interference What kind of interference will we get for a particular thickness? The wavelength of light in the film is equal to: 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 wavelength 2L = (m + ½) ( /n 2 ) -- The two rays will produce constructive interference if 2L is equal to a wavelength 2L = m ( /n 2 ) --
15
Interference 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
16
Color 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: = (2Ln) / (m + ½) = 1330 nm / (m + ½) = Only 532 nm is in the visible region and is green
17
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?
18
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
19
Path Length Difference Depends on: L and Example: double slit interference Equations: d sin = m -- maxima d sin = (m + ½) -- minima
20
Different Index of Refraction Depends on: L,, n Example: combine beams from two media Equations: N 2 - N 1 = (L/ )(n 2 -n 1 )
Similar presentations
© 2025 SlidePlayer.com. Inc.
All rights reserved.