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)
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) + +
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
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 …
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) tt E 2 (t) E0E0 E P (t) ERER /2 E0E0 tt E 1 (t) t+ E0E0 E 2 (t)
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
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.
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.
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
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
Chapter 34 – 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
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 ?
Solution Corresponding path length difference: Phase difference for going though plastic sheet: Angle of central max is approx: Thus the distance y’ is: gives
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
Examples : constructive: 2t = (m +1/2) n destructive: 2t = m n constructive: 2t = m n destructive: 2t = (m +1/2) n
Application Reducing Reflection in Optical Instruments
Experimental Observations: (pattern produced by a single slit ?)
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
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 ?
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.
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 ( ) /
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.
Fresnel Diffraction (or near-field) Incoming wave Lens Screen P (more complicated: not covered in this course)
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
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:
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
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
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!
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 !
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