Wave Optics Interference and other Mysteries Explained Particle or wave?
You Predict What happens when a pitcher throws lots of fastballs at two holes? What pattern do you see on the wall beyond?
Predict This What happens when a beam of light goes through two tiny slits? What pattern do you expect on the screen beyond?
Double Slit Interference Applet What if light from two slits interferes? ucation/Java/1998.Winter/Projects/pierce- woods/project/bin/projApp.htmhttp:// ucation/Java/1998.Winter/Projects/pierce- woods/project/bin/projApp.htm Experiment performed by Young in 1801 Convincing evidence for wave nature of light Another Double Slit Applet
Double Slit Derivation ics101/StudySheets/Double.htmlhttp://noether.physics.ubc.ca/Teaching/Phys ics101/StudySheets/Double.html
Understanding Path Difference dsin d is the path difference for waves traveling to a given point on the fringe pattern If path difference is an integer number of wavelength, interference is constructive (bright fringe) If path difference is a half integer number of wavelengths, interference is destructive (dark fringe)
Line Spacing d sin = m m constructive d sin m m destructive x = L Lm d for bright fringes using sin ~ in radians x L d ~ tan = x/L
What Will Happen to the Fringe Spacing… x = Lm d …if the wavelength increases? …if the distance to the screen increases? …if the slit width increases?
Find x = Lm d dx/Lm = 6.12 x 10 –7 m = 612 nm x = 2.5 mm L= 2.6 m d = 0.6 mm x 10 –7 m = 578 nm
Varying the Slit Separation Courtesy of Siltec Ltd.
x = Lm d xd/L for adjacent fringes= 4x10 -3 x 5x = 7.7 x = 770 nm
Problem When white light passes through two slits 0.50mm apart an interference pattern appears on a screen 2.5m away. The fringe separation is 2.5 mm for the violet light. Find the wavelength of this light. Hint: solve x = Lm d for L dx/mL = dx/L = 5.0 x m = 500nm
Single Slit Interference Also called diffraction Fringes are larger Size of fringes decreases out from center of pattern Derivation h25WO/Diff.html h25WO/Diff.html
Homework Ch 24 Problems 3,5,7,9,10(613 nm), 11
More On Single Slit Interference Pattern dominated by central maximum Called central diffraction maximum Width measured from minimum to minimum Twice as wide as other fringes Much brighter than other fringes
Single Slit Diffraction- Varying the Slit Width Fringes get bigger as slit gets smaller
Double Slit vs. Single Slit Double d sin = m bright d sin m Dark modest central maximum m=0,1,2,etc Single d sin = m dark d sin m bright Dominant central maximum Fringes usually larger m=1,2,3, etc
Single Slit Problem 500 nm light is incident on a slit of width mm which is 1.0 m from a screen. Find the width (2 of the central diffraction maximum. (hint: use dsin m degrees
Dispersion Spreading of white light into spectrum of wavelengths
Why Dispersion Occurs Index of refraction, n, depends on wavelength Typically n decreases as increases Exit angle from prism depends on White light
Forming a Rainbow Java Applet
Diffraction Gratings Diffraction Grating has thousands of lines per cm cut into glass plate Light from each slit interferes with light from all other slits Analysis like Double Slit Sin = m /d principal maxima (bright) Lines very close so maxima occur at large angles is maximum possible so only a few maxima (“orders”)exist
Courtesy
Grating Set-Up Courtesy A.Mercury lamp B.Spectrometer C.Gratings
How Grating Works in Practice Purpose is to analyze spectrum of light from source such as unknown gas or a star
Spectrum Produced by Grating
Diffraction Grating Problem A grating contains 5000 lines per cm –Find the slit separation (d) in meters 500 nm light is incident on the grating. –At what angles will the first, second and third orders be observed? Is there a fourth order? –Hint: use Sin = m /d 2 x m m= ; m= ; m= ; m= not visible = 0.5x10 -6 m/2x10 -6
Thin Film Interference Let film thickness = t Assume no net phase change on reflection 2t = m m Condition for bright 2t = (m+1/2) Condition for dark For one hard reflection 2t =m becomes condition for dark
Complication #1 Soft and Hard Reflections SOFT beam exits higher n material No phase shift on reflection HARD beam enters higher n material Phase shift of /2 on each reflection Air n=1.0 Coating n=1.38 Glass n=1.52
Complication #2 What Wavelength to Use Use n = n where n is the index of refraction in the medium where refraction occurs (the film) Example: green 500nm light leaves air and enters a thin layer of oil n=1.25 floating on water. What wavelength should be used to find the minimum thickness of oil for the oil to appear green? n = /1.25 = 500nm/1.25 = 400nm
Oil On Water Courtesty of How many hard reflections are there? What is the total phase shift due to reflections? What should you use? 1 n =
What Really Happens This shows why the wavelength must be divided by n in the film – to get the two reflected waves to interfere completely destructively
Anti-reflective Coating
Problem A coating of MgF on glass has an index of refraction of What is the minimum thickness of this coating to prevent reflection of 550 nm light? 2t =(m+1/2) n For m = 0 t = n /4 t= t = 100nm
Puzzle If the two reflected beams interfere destructively, what happens to their energy? It is transmitted! That’s the point of an antireflective coating.
Multi Coating Super-Multi-Coating (SMC) - Pentax's unique seven layer coating applied to each lens element of the lens to increase light transmission, increase color saturation and to help prevent flare.
Polarization Intensity varies as polarizer is rotated Courtesy Siltec Ltd.
Polarization Courtesy of 3M Corporation In linear Polarization light vibrations become confined to a single linear plane
Polarization Analogy
Two Polarizers What will happen if the polarized beam hits another polarizer rotated 90 0 from the first?
Propagation of a Linearly Polarized Electromagnetic Wave Animation Courtesy Siltec Ltd.
Courtesy: c.edu/courses/phys112/su mmer97/lectures/lect24/s ld014.htm
n = c/v
Link to Interference Applet In ripple tank link