Interference Applications Physics 202 Professor Lee Carkner Lecture 23

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) = 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 )

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

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  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

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

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

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

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

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

Next Time  Read:

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

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

What happens to the distance between the fringes if the distance between the slits increases? a)Increases b)Decreases c)Stays the same

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

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

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 )