2. Physics 1202: Lecture 26 Today’s Agenda Announcements: –Midterm 2: Friday Nov. 6… –Chap. 18, 19, 20, and 21 No HW for this week (midterm)No HW for this week (midterm) Optics –Interference –Diffraction

3. Interference

4. 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)

5. 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 … (destructive at m= ±1/2, ±3/2 …

6. Change of Phase Due to Reflection Lloyd’s mirror P2P2 P1P1 S I L Mirror The reflected ray (red) can be considered as an original from the image source at point I. Thus we can think of an arrangement S and I as a double-slit source separated by the distance between points S and I. An interference pattern for this experimental setting is really observed ….. but dark and bright fringes are reversed in order This mean that the sources S and I are different in phase by 180 0 An electromagnetic wave undergoes a phase change by 180 0 upon reflecting from the medium that has a higher index of refraction than that one in which the wave is traveling.

7. Change of Phase Due to Reflection 180 0 phase change n1n1 n2n2 n 1 <n 2 n1n1 n2n2 no phase change n 1 >n 2

8. Interference in Thin Films Air Film t 1 2 180 0 phase change no phase change A wave traveling from air toward film undergoes 180 0 phase change upon reflection. The wavelength of light n in the medium with refraction index n is The ray 1 is 180 0 out of phase with ray 2 which is equivalent to a path difference n /2. The ray 2 also travels extra distance 2t. Constructive interference Destructive interference

9. Chapter 26 – Act 1 Estimate minimum thickness of a soap-bubble film (n=1.33) that results in constructive interference in the reflected light if the film is Illuminated by light with =600nm. A) 113nm B) 250nm C) 339nm

10. Problem Consider the double-slit arrangement shown in Figure below, where the slit separation is d and the slit to screen distance is L. A sheet of transparent plastic having an index of refraction n and thickness t is placed over the upper slit. As a result, the central maximum of the interference pattern moves upward a distance y’. Find y’ where will the central maximum be now ?

11. Solution Corresponding path length difference: Phase difference for going though plastic sheet: Angle of central max is approx: Thus the distance y’ is: gives

12. Phase Change upon Reflection from a Surface/Interface Reflection from Optically Denser Medium (larger n) Reflection from Optically Lighter Medium (smaller n) by analogy to reflection of traveling wave in mechanics 180 o Phase ChangeNo Phase Change

13. Examples : constructive: 2t = (m +1/2) n destructive: 2t = m n constructive: 2t = m n destructive: 2t = (m +1/2) n

14. Application Reducing Reflection in Optical Instruments

15. Diffraction

16. Experimental Observations: (pattern produced by a single slit ?)

17. First Destructive Interference: (a/2) sin  = ± /2 sin  = ± /a m th Destructive Interference: (a/4) sin  = ± /2 sin  = ± 2  /a Second Destructive Interference: sin  = ± m /a m=±1, ±2, … How do we understand this pattern ? See Huygen’s Principle

18. So we can calculate where the minima will be ! sin  = ± m /a m=±1, ±2, … Why is the central maximum so much stronger than the others ? So, when the slit becomes smaller the central maximum becomes ?

19. Phasor Description of Diffraction Can we calculate the intensity anywhere on diffraction pattern ?  =  =  N   / 2  =  y sin (  ) /  = N  = N 2   y sin (  ) / = 2  a sin (  ) / Let ’ s define phase difference (  ) between first and last ray (phasor) 1st min. 2nd max. central max. (a/  sin  = 1: 1st min.

20. Yes, using Phasors ! Let take some arbitrary point on the diffraction pattern This point can be defined by angle  or by phase difference between first and last ray (phasor)  The arc length E o is given by : E o = R  sin (  /2) = E R / 2R The resultant electric field magnitude E R is given (from the figure) by : E R = 2R sin (  /2) = 2 (E o /  ) sin (  /2) = E o [ sin (  /2) / (  /2) ] I = I max [ sin (  /2) / (  /2) ] 2 So, the intensity anywhere on the pattern :  = 2  a sin (  ) /

21. Other Examples What type of an object would create a diffraction pattern shown on the left, when positioned midway between screen and light source ? A penny, … Note the bright spot at the center. Light from a small source passes by the edge of an opaque object and continues on to a screen. A diffraction pattern consisting of bright and dark fringes appears on the screen in the region above the edge of the object.

22. Fraunhofer Diffraction (or far-field) Incoming wave Lens  Screen

23. Fresnel Diffraction (or near-field) Incoming wave Lens Screen P (more complicated: not covered in this course)

24. Resolution (single-slit aperture) Rayleigh’s criterion: two images are just resolved WHEN: When central maximum of one image falls on the first minimum of another image sin  =  / a  min ~  / a

25. Diffraction patterns of two point sources for various angular separation of the sources Resolution (circular aperture)  min = 1.22 (  / a) Rayleigh’s criterion for circular aperture:

26. EXAMPLE A ruby laser beam ( = 694.3 nm) is sent outwards from a 2.7- m diameter telescope to the moon, 384 000 km away. What is the radius of the big red spot on the moon? a. 500 m b. 250 m c. 120 m d. 1.0 km e. 2.7 km  min = 1.22 (  / a) R / 3.84 10 8 = 1.22 [ 6.943 10 -7 / 2.7 ] R = 120 m ! Earth Moon

27. Two-Slit Interference Pattern with a Finite Slit Size I diff = I max [ sin (  /2) / (  /2) ] 2 Diffraction ( “ envelope ” function):  = 2  a sin (  ) / I tot = I inter. I diff Interference (interference fringes): I inter = I max [cos (  d sin  /  ] 2 smaller separation between slits => ? smaller slit size => ? The combined effects of two-slit and single-slit interference. This is the pattern produced when 650-nm light waves pass through two 3.0- mm slits that are 18 mm apart. Animation

28. Example The centers of two slits of width a are a distance d apart. Is it possible that the first minimum of the interference pattern occurs at the location of the first minimum of the diffraction pattern for light of wavelength  ? d a a 1st minimum interference: d sin  = /2 1st minimum diffraction: a sin  = The same place (same  ) : /2d = /a a /d =  No!

29. Application X-ray Diffraction by crystals Can we determine the atomic structure of the crystals, like proteins, by analyzing X-ray diffraction patters like one shown ? A Laue pattern of the enzyme Rubisco, produced with a wide-band x-ray spectrum. This enzyme is present in plants and takes part in the process of photosynthesis. Yes in principle: this is like the problem of determining the slit separation (d) and slit size (a) from the observed pattern, but much much more complicated !

30. Determining the atomic structure of crystals With X-ray Diffraction (basic principle) 2 d sin  = m  m = 1, 2,.. Crystalline structure of sodium chloride (NaCl). length of the cube edge is a = 0.562 nm. Crystals are made of regular arrays of atoms that effectively scatter X-ray Bragg ’ s Law Scattering (or interference) of two X-rays from the crystal planes made-up of atoms