1 W15D1: Interference and Diffraction Experiment 6 Today’s Reading Course Notes: Sections 14.4-14.9Announcements No Math Review Week 15 Tuesday PS 11 is only for practice. It will not be graded. Final Exam Practice Problems posted on ExamPrep page Final Exam Review Friday May 16 from 12 noon-2 pm in 26-152 Friday May 16 from 2 pm-4 pm in 26-152 Final Exam May 19 from 9 am -12 noon Second floor Johnson Athletic Center 23 Outline Review Interference Diffraction Interference and Diffraction Experiment 64 Review The Light Equivalent: Two Slits5 Young’s Double-Slit Experiment Bright Fringes: Constructive interference Dark Fringes: Destructive interference6 Double Slit Interference Source of Applet: http://online.cctt.org/physicslab/content/applets/doubleslit/slits.htm7 Two In-Phase Sources: Geometry Assuming L >> d : Extra path length δ= d sinθ( ) δ= d sinθ= mλ⇒ Constructiveδ= d sinθ= m +12( )λ⇒ Destructive8 Interference for Two Sources in Phase Constructive: yconst= mλL / d; m = 0, ± 1, ⋅ ⋅ ⋅Destructive: δ= (m + 1 / 2)λ ydest= (m + 1 / 2)λL / d; m = 0, ± 1, ⋅ ⋅ ⋅ δ= mλ Assume L >> d >>λ y = L tanθ≈ Lsinθ ⇒δ= d sinθ= dy L9 Discussion Question What is the relationship between phase shift and path length? Determine an analytic relationship between these two concepts.10 Phase Shift = Extra Path? δλ=φ2π=m constructivem +12destructive⎧⎨⎪⎩⎪What is exact relationship between extra path length and phase shift? sin(k(x +δ)) = sin(kx + kδ)= sin(kx +2πλδ) ≡ sin(kx +φ)11 Intensity Distribution E1= E0sinωt; E2= E0sin(ωt +φ) ETotal= E1+ E2= 2E0cos(φ/ 2)sin(ωt +φ/ 2) sin A + sin B = 2sin(( A + B) / 2)cos(( A − B) / 2) < I > ∝ ETotal2≡ I0cos2(φ/ 2) = I0cos2(πδ/λ)What is intensity of two waves out of phase? Average Intensity with phase shift : Use φ= 2πδ/λ12 Average Intensity < I > = I0cos2(πd sinθ/λ)Average intensity:13 Diffraction14 Diffraction Diffraction: The bending of waves as they pass by certain obstacles No spreading after passing though slits Spreading after passing though slits No Diffraction Diffraction15 Single-Slit Diffraction δ= r1− r3= r2− r4= (a / 2)sinθDestructive interference: a sinθ= mλm = ±1,± 2,...“Derivation” (Motivation) by Division: Divide slit into two portions: δ= (a / 2)sinθ=λ/ 2 ⇒ a sinθ=λDon’t get confused – this is DESTRUCTIVE! Now divide slit into four portions: δ= (a / 4)sinθ=λ/ 2 ⇒ a sinθ= 2λGeneralization:16 Intensity Distribution Destructive Interference: a sinθ= mλm = ± 1, ± 2,...17 Concept Question: Lower Limit? Using diffraction seems to be a useful technique for measuring the size of small objects. Is there a lower limit for the size of objects that can be measured this way? 1. Yes – and if we used blue light instead of red light we can measure even smaller objects than the ones we measure using red light 2. Yes – and if we used blue light instead of red light we couldn’t even measure objects as small as the ones we measure using red light 3. No18 Concept Q. Answer: Lower Limit? Once the feature size a is as small as the light wavelength you can’t go to an angle large enough to satisfy the above equation for any m > 0. Blue light has a shorter wavelength than red light, so you can measure smaller sizes using blue light. Answer: 1. we have the condition that There is a lower limit imposed by the condition, namely that ! sinθ= mλ/ asin 1 aθλ≤⇒ ≤19 Interference & Diffraction Together20 Two Slits With Finite Width With more than one slit having finite width a, we must consider 1. Diffraction due to the individual slit 2. Interference of waves from different slits21 Lecture Demonstration: Double Slits with Diffraction http://tsgphysics.mit.edu/front/?page=demo.php&letnum=P%2010&show=022 Two Slits With Finite Width 23 Interference & Diffraction24 Con. Q.: Interference & Diffraction The resulting pattern on a screen far away is shown above, with distantly-spaced zeroes of the envelope, as indicated by the length X above, and closely-spaced zeroes of the rapidly varying fringes, as indicated by the length Y above. Which length in the pattern above is due to the finite width a of the apertures? 1. X 2. Y 3. X and Y 4. Neither X nor Y Coherent monochromatic plane waves impinge on two long narrow apertures (width a) that are separated by a distance d with d > a.25 Concept Q. Ans.: Inter. & Diffraction Answer: 1. The ‘envelope’ length X depends on slit width. You could infer this in two ways. 1) Slit width a < slit separation d. Angles and size scale inversely, so the bigger features come from a. 2) Interference patterns are roughly equal in magnitude while diffraction creates a strong central peak. So the envelope is from diffraction.26 Worked Problem: Interference In an experiment you shine red laser light (600 nm) at a slide and see the following pattern on a screen placed 1 m away: a) Are you looking at a single slit or at two slits? b) What are the relevant lengths (width, separation if 2 slits)? What is the orientation of the slits? You measure the distance between successive fringes to be 20 mm27 Solution: Interference First translate the picture to a plot: (a) Must be two slits ad -80 -60 -40 -20 0 20 40 60 80IntensityHorizontal Location on Screen (mm)28 Solution: Interference y = L tanθ≈ Lsinθ= Lmλ/ d d = Lmλ/ y = 1m( )1( )600nm( )20mm( )= 1m( )6 × 10−7( )2 × 10−2( )= 3 × 10−5m( )( )sin 1sin 3adθ λθ λ==51310ada−⇒ == m-80 -60 -40 -20 0 20 40 60 80IntensityHorizontal Location on Screen (mm)At 60 mm…29 You just observed an interference pattern using a red laser. What if instead you had used a blue laser? In that case the interference maxima you just saw would be Concept Q.: Changing Wavelengths 1. closer together. 2. further apart. 3. came distance apart.30 Concept Q. Ans.: Changing Colors Blue light is a higher frequency (smaller wavelength) so the angular distance between maxima is smaller for blue light than for red light Answer: 1. Closer Together d sinθ= mλ31 Experiment 6, Part I: Measure Laser Wavelength yconst= mλL / d ; m = 0, ± 1, ± 2, ⋅ ⋅ ⋅32 Experiment 6, Part II: Interference from a CD Diffraction Grating d = distance between openings yconst= mλL / d; m = 0, ± 1, ⋅ ⋅ ⋅ d
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