Double-Slit Interference
- 1. LearningObject – CamilleAlvarez (LC2)
- 2. PhET Simulation • The points of constructive interference appear as bright bands (green) on the screen. • The points of destructive interference appear as dark bands. These bands are referred to as “fringes”. Let’s go into more detail.
- 3. We can determine the angles where maximum light intensity (bright fringes) occurs by thinking about a right triangle. Fraunhofer Condition: The distance from A1 to B is approximately the same as the distance from C to B ONLY IF the distance from the screen L is much greater than the slit separation d. The extra distance that light travels from A2 to the screen at B is Dsinθ This extra distance coincides to the path difference (Hence the fact that light travels different lengths from both slits to a point on the screen). http://theory.uwinnipeg.ca/ physics/light/img30.gif C A1 A2 BThese angles are the same dsinθ
- 4. Assuming the light waves from the source are in phase, the path difference for Constructive Interference: Δx = mλ Destructive Interference: Δx = (m+1/2)λ
- 5. Red = bright fringe (constructive interference) Blue = dark fringe (destructive interference)
- 6. When the path difference corresponds to exactly a whole number of λ, constructive interference (bright fringes) occurs. So: Dsinθ = mλ For all values of m other than zero that is an integer, the maxima for different wavelengths have different angles. An m value of zero means that θ = 0 ° and the light would appear in the middle of the screen. Path difference Δx !!
- 7. Accordingly, when the path difference corresponds to an odd number of half λ, destructive interference (the dark fringes) occurs. So: Dsinθ = (m+1/2)λ
- 9. Let’s assume that y is much less than L so we can use small angle approximation for sinθ ≈ y/L Therefore dsinθ = mλ becomes d(y/L) = mλ y = (mλL)/d This equation allows us to calculate the distance of a bright fringe to the point on the screen opposite of the center of the slits.
- 10. Red has the largest wavelength in the visible spectrum. Using y = (mλ)L/d we can see mathematically that the distance between bright fringes is a lot bigger than purple (smallest wavelength).