Today’s Lecture Interference Diffraction Gratings Electron Diffraction

Can we see any interference without a laser?

Some math: the slits are two coherent sources Some math: the slits are two coherent sources. The distances to the observation points are r1 and r2. Their difference for small angles q, and small y/L

Approximation used: for small Constructive (a bright strip) Destructive (a dark strip)

Constructive (a bright strip) Approximation used: Constructive (a bright strip) Consider two slits .075mm apart located 1.5m from screen. If the 3rd order bright fringe is 3.8cm from the screen center what is l?

In general, the distribution of intensity on the screen: − intensity of either wave alone Bright fringes: Dark fringes:

Multiple-Slit Interference and Diffraction Gratings From the diffraction limit with only two slits the intensity pattern is What if you require better resolution? Consider multiple slits. For multiple slits The location of the bright fringes still satisfies The location of the dark fringes now satisfy Here N is the number of slits and m is not an integer multiple of N. This means there are N-1 dark fringes between each bright fringe.

“Diffraction Grating” Multiple-Slit Interference and Diffraction Gratings More minima between bright fringes results in better resolution of the bright fringes. A set of very many closed spaced slits is called a “Diffraction Grating”

White Light and a Diffraction Grating This figure shows spectra that is observed when passing white light through a diffraction grating (note m=0). The orders of the diffraction, m, are separated vertically for presentation purposes. Note the increased separation between violet and red as the order of the dispersion increases.

Multiple-Slit Interference and Diffraction Gratings More minima between bright fringes results in better resolution of the bright fringes. Using a diffraction grating with 6000 slits/cm what is the angular separation between the hydrogen a spectral line (l=656.3nm) and the hydrogen b line (l=486.1nm)? The same criteria for the diffractions peaks applies, dsinq =ml. For this grating the slit spacing is d=(1/600)mm.

Resolving Power and Diffraction Gratings For a grating with N slits the criteria for the mth order maximum is The adjacent minimum occurs at Maximum for l’ = minimum for l

Multiple-Slit Interference and Diffraction Gratings A binary star system has one massive star at rest and another smaller star rotating about it. The hydrogen a line for the rotating star is Doppler shifted from l=656.272nm to l=656.215nm. What order spectrum is required for astronomers to resolve these lines when using a grating that is 2.5cm wide with 2000 slits/cm? Since the order must be an integer, the resolution of the two wavelengths is third order, m = 3.

X-Ray Diffraction When atomic spacing is comparable to x-ray wavelengths then the reflected beam satisfies Bragg condition: Bragg scattering is used to determine the positions of atoms in a crystal structure, or even the locations of atoms in organic molecules. Watson and Crick won the Nobel prize for X-Ray diffraction analysis of the “DNA molecule”

Interference in Thin Films Constructive Interference: Destructive Interference: Is the wavelength inside the film and

Newton’s Rings Interference in thin films is a very sensitive test for optical systems. Above you see the rings induced from the lens of a telescope.

Interference in Thin Films Constructive Interference: Destructive Interference: If the film were illuminated with 650nm light, how many bright bands in the film with n=4/3 will appear? There are 4 bright bands, why?? (b) What part of the film will be dark? Since the smallest wavelength for visible light is l=400nm,

No Interference with Bullets Bullets come out randomly. If you block slit 2 then the bullets have a probability distribution P1 with an analogous result when you block slit 1. If you open both slits then the bullets have the probability distribution P12 = P1 + P2

Interference with Waves With water waves (or monochromatic light), If you block slit 2 then the intensity, I1, is given by intensity distribution |h1|2 with an analogous result when you block slit 1. If you open both slits then the interference results in an intensity distribution I12 = |h1 + h2|2 NOT |h1|2+|h2|2.

Interference of Electron Waves? Electrons of identical energy come out randomly. If, (b), you block slit 2 then the electrons have a probability distribution P1=|f1|2. There is an analogous result when you block slit 1. If, (c), you open both slits then the electrons have the probability distribution P12 = P1 + P2?? NO, P12=|f1+f2|2. The detector only measures whole discrete electrons! but with the probability of a wave!

Interference of Electron Waves with a Light Detector If the light from the source interacts with the electrons then you should observe a flash through one of the slits. But then the interference pattern Disappears! So are the electrons particles or waves??

Interference of Electron Waves with a Light Detector + To answer this question, we turn down the intensity of the light source. Now the flashes occur less often, but are just as bright when they do occur. The resulting pattern is partially that of particles and waves. Are the electrons somehow interfering with each other? NO! The same interference pattern results even when electrons arrive one at a time, but it takes longer to develop. www.hqrd.hitachi.co.jp/em/doubleslit.cfm

Interference of Electron Waves with a Light Detector Remembering the diffraction limit for two slits, what happens when we use light with ever increasing wavelengths? When l approximately equals d then a fuzzy flash appears to come through both slits, and ?? The interference pattern returns! It seems that if you measure a particle property of the electron, it behaves as a particle, but it you measure a wave property it behaves as a wave!

Interference of Electron Waves with a Light Detector DeBroglie wavelength is defined as p=h/l or l=h/p. In this expression h is Planck’s constant, h=6.6x10-34Jsec. This expression holds for both light particles (photons) as well as any particle with mass! If the momentum of the photon is large enough, when it scatters off of an electron, it destroys the interference pattern.

That’s All Folks! This completes Physics 2C. I hope you enjoyed at least some of it! There will be a review session tomorrow that should be helpful in preparing for the final! Good Luck to Everyone!