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Physics 1502: Lecture 32 Today’s Agenda

Announcements: Midterm 2: graded after Thanks Giving Homework 09: Friday December 4 Optics Eye interference. Physics 1502: Lecture 32 Today’s Agenda. The EYE. I 2. I 1. ~f e. ~f o. L. eyepiece. objective. Objective (f ob < 1cm). Eyepiece (f eye ~5cm). L. f ob. f eye. o 1.

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Physics 1502: Lecture 32 Today’s Agenda

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  1. Announcements: Midterm 2: graded after Thanks Giving Homework 09: Friday December 4 Optics Eye interference Physics 1502: Lecture 32Today’s Agenda

  2. The EYE I2 I1 ~fe ~fo L eyepiece objective

  3. Objective (fob< 1cm) Eyepiece (feye~5cm) L fob feye o1 i1 I1 h h1 O I2 h2 Magnification: Compound Microscope

  4. Objective (fob~ 250cm) Eyepiece (feye~5cm) fob feye i1 I1     h1 h2 I2 Angular Magnification: Refracting Telescope Star

  5. Interference

  6. What happens when two waves collide ? They add point by point Superposition Why? Because the wave equation is linear. This is the principle of superposition.

  7. If you added the two sinusoidal waves shown in the top plot, what would the result look like ? Lecture 32 – Act 1

  8. Wavefronts: slit acts like point source Rays A wave through a slit

  9. A wave through two slits (two coherent point sources)

  10. Intensity What happens when two light waves are present at the same point in space and time? What will we see? Intensity! Add Amplitudes! (electric fields or magnetic fields) Brightness ~ <Amplitude2> ~ ½ E02

  11. Suppose laser light of wavelength l is incident on the two-slit apparatus as shown below. Lecture 32 – Act 2 Which of the following statements are true? (A) There are new patterns of light and dark. (B) The light at all points on the screen is increased (compared to one slit). (C) The light at all points on the screen is decreased (compareed to two slits).

  12. q q A wave through two slits d L Screen Assume L is large, Rays are parallel

  13. d DP=d sinq A wave through two slits In Phase, i.e. Maxima when DP = d sinq = nl Out of Phase, i.e. Minima when DP = d sinq = (n+1/2)l q Screen

  14. + + A wave through two slits In Phase, i.e. Maxima when DP = d sinq = nl Out of Phase, i.e. Minima when DP = d sinq = (n+1/2)l

  15. Note that you could derive the reflectance equation(qi=qR) using a particle model for light. Bouncing balls. You could also derive Snell’s Law for particles. n1sin (qi)=n2sin(q2) The particles change speed in different media (Newton did just this) You cannot get a particle model for these interference effects. You would have to magically create particles at the bright spots and annihilate them at the dark spots. Interference effects mean that light must be made up of waves. Waves and Interference

  16. What determines the wave amplitude at P? The difference in the path lengths! (ie d= S1P - S2P) If the d is an integral number of wavelengths, the phase difference is zero and we get constructive interference. If d is l/2, 3 l/2, 5 l/2, etc, we get destructive interference. The general case is given by: The amplitude for the wave coming from S1: The amplitude for the wave coming from S2: The amplitude for the total wave at P : with The Amplitudes

  17. What is the intensity at P? The only term with a t dependence is sin2( ).That term averages to ½ . If we had only had one slit, the intensity would have been, So we can rewrite the total intensity as, with The Intensity

  18. We can rewrite intensity at point P in terms of distance y The Intensity Using this relation, we can rewrite expression for the intensity at point P as function of y Constructive interference occurs at where m=+/-1, +/-2 …

  19. Note that the angle between bright spots is given by, sinq = nl/d To see effect we want l  d. l>d means no bright spot, l<<d means bright spots too close together. For an x-ray, l = 1 Å, E = 10 keV  d ~ 1-20 Å. Interference patterns off of crystals For an electron, l = 1 Å, E = 100 meV (deBroglie - 1925) 100 meV means an electron is accelerated through a voltage of 0.1 V Interference patterns off of crystals Davisson and Germer, (1927) So, electrons are waves ?? d spacing

  20. E0 E2(t) wt+f E0 E1(t) wt E0 E2(t) f/2 f ER EP(t) E1(t) E0 wt Consider a sinusoidal wave whose electric field component is Phasor Addition of Waves Consider second sinusoidal wave The projection of sum of two phasors EPis equal to

  21. ER=2E0 E0 ER 900 ER 450 E0 E0 E0 E0 E0 ER=0 E0 2700 ER=2E0 E0 E0 E0 ER E0 E0 Phasor Diagrams for TwoCoherent Sources

  22. SUMMARY 2 slits interference pattern (Young’s experiment) How would pattern be changed if we add one or more slits ? (assuming the same slit separation ) 3 slits, 4 slits, 5 slits, etc.

  23. Phasor: 1 vector represents 1 traveling wave 2 wave interference single traveling wave

  24. N-slits Interference Patterns F=0 F=90 F=180 F=270 F=360 N=2 N=3 N=4

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