In double-slit interference a series of bright and dark lines are formed. For vertical slits, the light spreads out horizontally on either side of the incident beam into a pattern called interference fringes (see figure below). The bright fringes are formed at the height ym where the two waves interference is pure constructive, hence the following condition is satisfied d sin(0) = m2, m = 0,+1,+2, . Thus, the distance from the central maximum to the mth bright fringe will be mAD Ym = d Where D is the distance between the slit and the screen and 2 is the wavelength of light. This approximation is accurate enough as long as 0 is small thus sine s tane - D The closer the slits are (smaller d), the more the bright fringes spread apart (bigger Aym). Since the ym also depends on wavelength, 2, thus the maximums don't occur in the same place for different wavelengths. 02 Experiment 2: Double Slit Interference Using the following simulation of the double slit interference, we want to investigate the dependency of the ym on both wavelength and slit size. https://ophysics.com/15.html Double Slit Interference 4 um Animate Pause Reset -3 um V Show Maxima V Show Scale 2 um V Show Wavefronts 1 um Sit Distance ( 1-3 um) 3 1 um Wavelength (400 - 700 nm) 400 2 um Show Interference Pattern 3 um 4 ur L= 10 um Question 3: [24 points, 1 for each] Change the wavelength and the slit distance following the instruction the table below. Use "Show scale" and measure the y, and y2. Record height of the first and second maximum in the table. Then answer the questions. Case Slit distance, Distance to screen, Wavelength, 2 [um] 400 Y1 [um] Yz[um] [un] p 10 L[um] 1 2.2 2 400 2.7 10 400 3.0 10 4 450 2.2 10 5 450 2.7 10 6. 450 3.0 10 7 500 2.2 10 500 2.7 10 9. 500 3.0 10 10 550 2.2 10 11 550 2.7 10 12 550 3.0 10 Question 4: [6 points, 3 points each] Does the change of slit distance or wavelength effects the height of the first and second maximum? Explain the dependency?