PH 103 Dr. Cecilia Vogel Lecture 8

Review Outline  diffraction  interference  coherence  Diffraction/interference examples  double - slit and diffraction grating  single slit  thin films  holograms

Two-slit interference  AKA Young’s experiment  Two waves start out in phase, but one travels farther  one wave gets behind (analogy: cars)  Geometry: slits  Observation screen 

Two-slit interference  Geometry if slits d and << L  difference in distance traveled =  d sin   or dy / L

Two-slit interference  Constructive interference if difference in distance traveled = integer # of wavelengths  d sin  = m  or dy / L = m  Destructive interference if difference in distance traveled = (integer- 1/2 )wavelengths  d sin  = (m - ½)  or dy / L = (m - ½)

How does interference pattern depend on  slit separation?  Larger d, smaller  y -- fringes closer  wavelength?  Longer, larger  y -- fringes farther  longer wavelengths diffract more  interference can tell yellow from red +green  material? = o /n, shorter  Bright fringes: dy/L = m  distance between fringes:  d  y/L = Two-slit interference

Many-slits = diffraction grating  Each pair of slits behaves like double-slit  Constructive interference if  d sin  = m  or dy / L = m  Destructive interference if  d sin  = (m - ½)  or dy / L = (m - ½)  Fringes are in same place as double-slit, but sharper

Many-slits = diffraction grating  How far apart are the slits?  Suppose the are 10 lines/cm, then there is one line in 1/10 cm = 0.1 cm  the lines are 0.1 cm apart  generally d = 1/(number of lines per unit length)

Single Slit diffraction pattern  Dark Fringes occur at (m =integer again)  So, width of center bright spot,  center  How does pattern depend on  slit width  narrower slit causes more diffraction & wider pattern  wavelength  longer wavelengths diffract more

One-slit/Two-slit Confusion  Note: W =slit width, but d=slit separation  Is the spot bright or dark?  Single-slit eqn is for dark (destructive interference)  Double-slit eqn is for bright (constructive interference)  What values of m are possible?  M = all integers for double slit  m = integers except zero for single slit Single-slitDouble-slit

Circular opening diffraction pattern  Circular diffraction pattern depends on  Aperture radius, a  smaller - more diffraction & wider pattern  wavelength  longer wavelengths diffract more  So two objects separated by ½ that can be resolved as separate bright spots = “Rayleigh Criterion”

Hologram  Light wave split in two by half-silvered mirror  One part shines on object, then reflects to film = “ object beam“  one part goes directly to film = “reference beam“  Interference of these two waves depends on how much further one travels than the other  3-D shape of the object recorded as  Developed film is like a complicated grating  When light falls on film, reproduces light from original object

One more thing about Reflection  Wave will change phase by 180 o (i.e. ½ ) if it reflects from material where the wave goes slower (higher n)  Wave will not change phase if it reflects from material where the wave goes faster (lower n) Demo

VERY Thin film  When light strikes a thin film at small incident angle, some reflects from top surface, some from bottom surface.  If the film is very thin, the only difference between the two reflected waves is if one has a phase change when it reflects.  Does each wave experience a phase change?

VERY Thin Films  Constructive interference if both or neither change phase  light comes from the smallest n, film is intermediate  or light comes from largest n, film is intermediate  Destructive interference if  film is largest n  or film is the smallest n

Thin Films  The two waves also differ in distance traveled by twice the thickness = 2t  If difference in distance traveled is integer number of cycles, then  same result as very thin film  beware: wavelength within film (use n of film)  2t = m /n  If difference in distance traveled = integer number of cycles + ½ cycle, then  opposite result from very thin film  again beware: wavelength within film  2t = (m+½) /n  Must know what happens to very thin film 1 st !

EXAMPLE Antireflective coating: Thin coating of material with n=1.25 on glass (n=1.55) makes 525-nm green light not reflect. How thick should the coating be? 1st: What happens when light from air hits a very thin film like this? n of film is intermediate, so it would be bright. 2nd: To make it dark instead (opposite), it must travel through a thickness given by: 2t = (m-½) /n t = (m-½) /2n=(½)(525nm)/(2(1.25)) = 105 nm =1050 Å