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

3. ~f e I1I1 eyepiece I2I2 ~f o objective L The EYE

6. Compound Microscope o1o1 h O I2I2 h2h2 f eye h1h1 I1I1 i1i1 Objective (f ob < 1cm) f ob L Eyepiece (f eye ~5cm) Magnification:

7. Refracting Telescope Star f eye I2I2 h2h2 f ob Objective (f ob ~ 250cm) Eyepiece (f eye ~5cm) i1i1 I1I1 h1h1 Angular Magnification:    

8. Interference

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

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

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

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

13. 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 ~ ~ ½ E 0 2

14. Lecture 32 – Act 2 Suppose laser light of wavelength  is incident on the two-slit apparatus as shown below. 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).

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

16. A wave through two slits Screen   P=d sin  d In Phase, i.e. Maxima when  P = d sin  = n Out of Phase, i.e. Minima when  P = d sin  = (n+1/2)

17. A wave through two slits In Phase, i.e. Maxima when  P = d sin  = n Out of Phase, i.e. Minima when  P = d sin  = (n+1/2) + +

18. Waves and Interference Note that you could derive the reflectance equation (  i =  R ) using a particle model for light. Bouncing balls. You could also derive Snell’s Law for particles. n 1 sin (  i )=n 2 sin(  2 ) 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.

19. The Amplitudes What determines the wave amplitude at P? The difference in the path lengths! (ie  = S 1 P - S 2 P) If the  is an integral number of wavelengths, the phase difference is zero and we get constructive interference. If  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 S 1 : The amplitude for the wave coming from S 2 : The amplitude for the total wave at P : with

20. The Intensity What is the intensity at P? The only term with a t dependence is sin 2 ( ).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

21. The Intensity We can rewrite intensity at point P in terms of distance y 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 …

22. d spacing Note that the angle between bright spots is given by, –sin  = n /d To see effect we want  d. >d means no bright spot, <<d means bright spots too close together. For an x-ray, = 1 Å, E = 10 keV  d ~ 1-20 Å. –Interference patterns off of crystals For an electron, = 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 ??

23. Phasor Addition of Waves Consider a sinusoidal wave whose electric field component is Consider second sinusoidal wave The projection of sum of two phasors E P is equal to E0E0 E 1 (t) tt E 2 (t) E0E0  E P (t) ERER  /2 E0E0 tt E 1 (t)  t+  E0E0 E 2 (t)

24. Phasor Diagrams for Two Coherent Sources E R =2E 0 E0E0 E0E0 E0E0 E0E0 ERER 45 0 E0E0 E0E0 ERER 90 0 E R =0 E0E0 E0E0 E0E0 E0E0 ERER 270 0 E R =2E 0 E0E0 E0E0

25. 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.

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

27. N=2N=4N=3      N-slits Interference Patterns