4) The photographer behind the website \"Pinhole Moustache\" constructed the following set up, and made the following image: Figure 3: The photographer from the website \"Pinhole Moustache\" experiments with light techniques a) Of the three types of slits we learnt about, which one is a pinhole camera most similar to, and why? b) From the photos above, make assumptions about the wavelength, how far the camera is from the wall, and how wide the various parts of the diffraction and interference pattern are. From these assumptions, calculate the width of the pinhole using formulas from our formula sheet. Note: The distance from the slit to the wall is not pictured, but make a best guess from the apartment set up. He also didn't need distance between the laser and the slit, just the slit and the wall. | am not sure why he did that. Wave Nature of Light sin 0= sin 0 =n- Ax w N d L f= n=1,2,3... n=1,2, 3... (m +_ ) a mA sin 0, V 7 1 2 sin 0.= n2 sin On = m d Ay _ sin 02 V 2 2 2 n1 W L W m=0,1, 2,3... m = 1,2,3, ... t= t= 2 4 n= 1, 2, 3... n= 1, 3, 5...To calculate the width of the pinhole using the given formulas, we need to make some assumptions about the wavelength (), the distance from the pinhole to the wall (L), and the width of the diffraction pattern (Ax or Ay). Let's assume: 1. Wavelength (\\): Assume the wavelength of the light used is 650 nm (nanometers), which is typical for a red laser. In millimeters, this is: A = 650nm = 650 x 10 \"mm = 0.65 x 10 *mm 2. Distance from the pinhole to the wall (L): Assume the distance is 2 meters (2000 mm). 3. Width of the central maximum (Ax or Ay): Assume the width of the central bright fringe (or the distance between the first minima on either side of the central maximum) is 10 cm (100 mm). Now, we use the formula: where: . Ax is the width of the central maximum (100 mm), . L is the distance from the pinhole to the wall (2000 mm), . A is the wavelength of the light (0.65 \\times 10^{-3} mm), . d is the width of the pinhole (which we need to find). Rearrange the formula to solve for d: L . X d = AxSubstitute the known values: g 2000 mm x 0.65 x 10~ mm B 100 mm Perform the calculation: 2000 x 0.65 x 10? - 100 1.3 100 d=0.013mm d d =13 pm Therefore, the width of the pinhole is approximately 13 micrometers (um). This result aligns with typical pinhole sizes used in pinhole cameras, which are usually in the range of tens of micrometers