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Young’s double slit experiment by Stanley ye


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physics 101 section 203, stanley ye

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  • Light and its nature have caused a lot of ink to flow during these last decades. Its dual behavior is partly explained by (1)Double-slit experiment of Thomas Young - who represents the photon’s motion as a wave - and also by (2)the Photoelectric effect in which the photon is considered as a particle. A Revolution: SALEH THEORY solves this ambiguity and this difficulty presenting a three-dimensional trajectory for the photon's motion and a new formula to calculate its energy. More information on :
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Young’s double slit experiment by Stanley ye

  1. 1. Young’s Double Slit Experiment BY STANLEY YE SECTION 203 PHYSICS 101
  2. 2. Historical Background In 1801, Thomas Young (1773-1829) performed an extraordinary experiment which would then later serve as evidence for the wave model of light. He had proposed that light also exhibited properties of a wave, which did not agree with the views of many other physicists during the early nineteenth century whom believed that light consisted of only particles.
  3. 3. Young’s Double Slit Experiment Purpose? - To demonstrate that light also had wave properties How was the experiment performed? - Two narrow slits were illuminated by a light source, which would then project an image onto a screen behind the slits.
  4. 4. What if light were only particles? If light were particles (did not exhibit properties of waves), then we would see two maximas that correspond to only rays that hit directly where the slits are located.
  5. 5. What if light had wave properties? If light had wave properties, then we would observe diffraction as we pass through the slits. We would then be able to see an interference patterns (constructive and destructive) from the waves as they meet each other. An image of alternating dark and light bands (interference fringes) would then appear on the screen. In fact, Young did observe an image of alternating dark and light bands, which confirms that light has wave properties.
  6. 6. Deriving a mathematical relationship Deriving an equation that determines at what angles, the maximas (light bands) occur. We can make a right-angle triangle that separates the distance from 2 to 3 into two parts. The distance from 1 to 3 is approximately equal to the distance from 4 to 3 (assuming that distance D >> d (slit distance)) according to the Fraunhofer condition. Using geometry, θ at point 1 is the same θ near point 4. Then the distance from point 2 to 3 and from 1 to 3 is a difference of d sinθ. When this difference corresponds to an integer, we observe constructive interference at angle θ. Hence, constructive interference occurs when d sinθ = mλ
  7. 7. Deriving a mathematical relationship Constructive interference occurs when d sinθ = mλ where m is an integer (spectral order) +/-1, +/- 2, +/3 .. Where it is positive on one side and the negative on the other. When m=0 (called the zero order) , no matter what the wavelength, θ is still 0. For other spectral orders, the maxima occurs at difference angles, depending on the wavelength. As wavelengths increase, so does the deflection angle. m values, longer wavelengths. Destructive interference occurs when d sinθ = (m + ½) λ Note: Ydark = λD(m + ½)/d
  8. 8. Deriving a mathematical relationship Finding displacement distance y (distance from the zero order to some spectral order): tan θ = y/D sin θ ≈ y/D Maximum spectral order: set sin θ = 1 and then round down m to the closest whole number using d sinθ = mλ
  9. 9. Check your understanding 1. Two screens are placed 10.3 m from each other. A white band of second order is seen on the screen 0.02m from zeroth order bright band. Find the wavelength of the light source if two slits are 0.05 cm apart. Approximately what colour is it? (D = 10.3 m as seen on previous diagram/slide) 2. If a light source of 700 nm illuminates the double slits (0.01 cm apart), at what angle would you observe the third order bright fringe?
  10. 10. Solutions 1. Looking at the given numbers, we can then determine m = 2 because of the second order fringe D (distance between the screens) = 10.3 meters y (distance from the central order to the spectral order) = 0.02m d (distance between the two slits) = 0.0005m λ = (y*d) / (m*D) λ = (0.02*0.0005) / (2*10.3) = 4.85e-7, which is 485 nm corresponding to approximately the wavelength of blue
  11. 11. Solutions 2. From the given variables: m = 3 because of the third order fringe λ = 700E-9 d (distance between the two slits) = 0.0001m d sinθ = mλ sinθ = mλ / d θ = arcsin (mλ / d) θ = arcsin (3*700E-9 / 0.0001) θ = 1.20 degrees
  12. 12. References ◦ "Lesson 58: Young's Double Slit Experiment." Lesson 58: Young's Double Slit Experiment. N.p., n.d. Web. 21 Mar. 2015.