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    Compare to L = 2.6X10 -5 m   6.5 is total destructive interference and so the above situation is brighter

Orders  When we look at a interference pattern on a screen each bright or dark spot is represented by a value of m called an order   The orders are symmetric e.g. the 5th order maxima is located both to the left and the right of the center at the same distance 

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  ) /  

Intensity  The intensity can be found from the electric field vector E: I  E 2  Where I 0 is the intensity of the direct light from one slit and  is the phase difference in radians  For any given point on the screen we can find the intensity if we know ,d, and I 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   If light is incident on a material with higher n, the phase shift is 0.5 wavelength   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 zero   Interference is due only to path length difference  Example:  If the thin film is in air, the first shift is zero and the second is 0.5   Have to add 0.5 wavelength shift to effects of path length difference  Example:

Path Length and Thin Films  For light incident on a thin film the light is reflected once off of the top and once off of the bottom   If the light is incident nearly straight on (perpendicular to the surface) the path length difference is 2 times the thickness or 2L 

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, the two reflected rays are in phase and they will produce destructive interference if 2L is equal to 1/2 a wavelength  The two rays will produce constructive interference if 2L is equal to a wavelength

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) = 2660 nm, 887 nm, 532 nm, 380 nm …   Real soap bubbles change thickness due to turbulence and gravity and so the colors shift

Interferometer  To get very accurate measurements of wavelength we use an interferometer   The beam is sent through a beam splitter, bounces off mirrors and is recombined to produce fringe patterns   1/2 movement of mirror will shift pattern by one fringe 

Interferometer

Interference: Summary  Interference occurs when light beams that are out of phase combine   The type of interference can depend on the wavelength, the path length difference, or the index of refraction 

Reflection  Depends on:  Example:  Equations:

Path Length Difference  Depends on:  Example:  Equations:   d sin  = (m + ½) -- minima

Different Index of Refraction  Depends on:  Example:  Equations: 