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- 1. TEACHING PRESENTATION IN PHYSICS Philippe MANUEL English - MASTER 1 EFTIS 2012 -203
- 2. The Double Slit Interference Experiment •Introduction and history •Wave interference properties •The principle of superposition •Thomas Young Experiment details •Conditions for constructive/desctructive interferences •Geometric statements about the interferences •Exercise : calculation of a light wavelength
- 3. The Double Slit Interference Experiment Conducted by Thomas Young in 1803 Showed an interference pattern A breakthrough of the light corpuscular theory of Isaac NEWTON Measured the wavelength of the light
- 4. Two Waves Interfering : Water waves example waves characteristics : Antinode <=> Minimal disturbance Node <=> Maximum disturbance
- 5. Principle of Superposition When two waves coincide, their displacements are added vectorially. + = This process is called reinforcement or constructive interference The two waves are in phase or have a phase difference of 2π or λ
- 6. Principle of Superposition + = This process is known as Annulment ( Flat wave ) or destructive interference The waves are out of phase or have a phase difference of π or λ/2.
- 7. Young’s experiment details Light source S1 S2
- 8. Double Slit Interference Experiment Diffraction step 1 Light source S1 S2
- 9. Double Slit Interference Experiment Diffraction step 2 S1 S2
- 10. c tive stru De Dark fringe s1 s2 Bright fringe Constructive
- 11. Double Slit Interference Experiment Typical Question How can we determine the conditions of constructive or destructive interference ( bright fringe versus back fringe ) ?
- 12. Bright fringes occur when the difference in path ∆p is an integral multiple of one wave length λ. p1 λ λ λ p2 p3 p4 Path difference Bright fringes: ∆p = 0, λ , 2λ, 3λ, … ∆p = nλ, n = 0, 1, 2, . . .
- 13. Dark fringes occur when the difference in path ∆p is an odd multiple of one-half of a wave length λ/2. λ λ p1 2 ∆p = n p2 2 λ n = odd p3 λ n= p3 1,3,5 … λ Dark fringes: ∆p = n n = 1, 3, 5, 7, . . . 2
- 14. Conditions required : ∆L=nλ X Géometric statements Assume L>> d S1 θ is very small ! θ d θ ∆L=d*θ=nλ X=L*θ S2 L -> nλ = d*X/L For n=0,1,2,3,…
- 15. Double Slit Interferences conditions Summary Location of fringes ( x) : x = nλ ∗ L/d ∆x = λ ∗ L/d Constructive (brights) n=0,1,2,3,….. Destructive (darks) n=1/2, 3/2, 5/2,….. ( m +1/2) ( odd numbers)
- 16. Application of wave length measurement exercise Monochromatic light from a single slit illuminates two narrow parallel slits. The centres of the two slits are 0.800mm apart. An interference pattern forms on a screen 50.0cm away. The fringe separation on the screen is 0.304mm. Find the wavelength λ of the light.
- 17. Exercise - Answers d = 0.800mm = 8.00 x 10-4m L = 50.0cm = 5.00 x 10-1m ∆x = 0.304mm = 3.04 x 10-4m λL ∆x = d d∆x λ= L 8.00 × 10 − 4 × 3.04 × 10 −4 = 5.00 × 10 −1 = 4.864 ×10 −7 λ = 4.86 ×10 −7 m
Editor's Notes
- Double slit experiment , also known as the Young's experiment, was conducted by english scoentist Thomas Young in the year 1803. This experiment confirmed the wave nature of light by showing light undergoes both diffraction and interference property of a wave , which showes that light behaves as a wave, not as bundles of light corpuscule as described into the light corpuscular theory of Isaac Newton in 1704 , from which Isaac Newton proposed that light consists of a stream of small particles travelling into spaces in straight lines.
- In Young’s experiment, light from a monochromatic ( coherent with a single frequency) source falls on two slits, setting up an interference pattern analogous to that with water waves. The Double Slit Experiment (Young's Interferometer) demonstrates interference of waves travelling in two dimensions using light waves .
- From a unique light source we generate a plane wave light source through a convergent optical lense. At the first slit level, we observe a diffraction phenomena where the plane wave of light is bended into circular waves as it passes through a narrow slit. S ingle slit is small enough to diffract light. This causes the same wave front of light to hit both slits S1 and S2
- By applying the principle of superposition explained before, we can deduce the following conditions : The waves must arrive to the point of study in phase. So their path difference must be integral multiples of the wavelength: = n n=0,1,2,3, There is a bright band at the centre of the pattern on the screen. This is the point where the path difference is zero . At this point the light from the two sources meets in phase . ie = 0 Other bright bands occur where the path difference is a whole number of wavelengths. At these points the light is also in phase. = m
- By applying the principle of superposition explained previously, we can deduce the following conditions : the waves must arrive to the point of study out of phase. So the path difference must be an odd multiple of /2 The dark bands on the screen occur where the light from the two slits is out of phase at that point on the screen. For these points, the path difference of light from the two sources is an odd number of half wavelengths. = (m + /2) = (2m +1 ) * /2
- To explain the origin of the interference pattern , consider the distance traveled from the two sources. At the center of the screen the waves from the two sources are in phase. As we move away from the center, the path traveled by the light from one source is larger than that traveled by the light from the other source. When the difference in path length is equal to a wavelength, constructive interference. Instead, when the difference is equal to half a wavelength, destructive interference occurs L very large comparatively to the slit spacing