# Interference Applications Physics 202 Professor Lee Carkner Lecture 25.

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Interference Applications Physics 202 Professor Lee Carkner Lecture 25

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 

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

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

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

Intensity Variation

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

Thin Film

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

Reflection Phase Change

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:

Path Length and Thin Films   What is the path length difference between the two reflected rays?   Don’t forget to include reflection shifts

Anti-reflective Coating

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 ) --

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 

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 

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?

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

Path Length Difference  Depends on: L and  Example: double slit interference  Equations:  d sin  = m -- maxima  d sin  = (m + ½) -- minima

Different Index of Refraction  Depends on: L,, n  Example: combine beams from two media  Equations:  N 2 - N 1 = (L/ )(n 2 -n 1 )