# Double-Slit Diffraction and Interference Let's break down the problem in two steps for each part, with explanations after each step. --- ## **Given Data** - **Wavelength, λ** = 632.8 nm (Helium-neon laser) - **Screen distance, L** = 2 meters - **Slit spacing, d** = .25 mm - **Goal:** a) Find the slit width, **a** b) Find the distances \( h_{12}, h_{13}, h_{14}, h_{15}, h_{1-2}, h_{1-3} \) --- ## **Step 1: Relate the Double-Slit Pattern** ### a) **Finding Slit Width \( a \)** The **envelope** (broad minima) in the double-slit pattern is due to **single-slit diffraction**. The minima occur at angles \(\theta\) given by: \[ a \sin\theta = m\lambda \] where \( m = \pm1, \pm2, \ldots \) For small angles (\(\sin\theta \approx \tan\theta \approx \frac{y}{L}\)), \[ a \frac{y}{L} = m\lambda \implies y = \frac{m\lambda L}{a} \] **Explanation:** This equation gives the position \(y\) of the minima due to the single-slit envelope. --- ### b) **Finding Positions of Interference Maxima** Interference maxima (bright fringes) for double slit are at: \[ d \sin\theta = n\lambda \] For small \(\theta\): \[ y = \frac{n\lambda L}{d} \] **Explanation:** This equation gives the position of the nth interference maximum due to the double-slit arrangement. --- ## **Step 2: Solve for the Requested Quantities** ### a) **Slit Width \( a \)** Assume you can measure the distance on the screen between the first minima on either side of the central maximum (i.e., from \(m = -1\) to \(m = +1\)), call this distance \(2y_1\): \[ 2y_1 = 2 \frac{\lambda L}{a} \implies a = \frac{2\lambda L}{2y_1} = \frac{\lambda L}{y_1} \] **Explanation:** If you measure (or are given) the distance between the first single-slit minima, plug in values for \(\lambda, L\), and \(y_1\) to solve for \(a\). --- ### b) **Distances Between Maxima (Fringe Spacing)** The distance between adjacent interference maxima (fringe spacing) is: \[ \Delta y = \frac{\lambda L}{d} \] The positions of the 1st, 2nd, ... maxima from the center are: \[ y_n = n\Delta y \] So, the distances: - \( h_{12} = y_2 - y_1 = \Delta y \) - \( h_{13} = y_3 - y_1 = 2\Delta y \) - etc. **Explanation:** The fringe spacing depends only on the wavelength, screen distance, and slit separation. --- ## **Final Answers** ### a) **Slit Width \( a \):** \[ a = \frac{\lambda L}{y_1} \] Where \(y_1\) is the distance from the central maximum to the first single-slit minimum. ### b) **Distances Between Maxima:** \[ \Delta y = \frac{\lambda L}{d} \] \[ h_{1n} = (n-1)\Delta y \] **Plug in values:** - \(\lambda = 632.8 \times 10^{-9}\) m - \(L = 2\) m - \(d = .25 \times 10^{-3}\) m \[ \Delta y = \frac{632.8 \times 10^{-9} \times 2}{.25 \times 10^{-3}} \approx .00506 \text{ m} = 5.06\, \text{mm} \] So: - \(h_{12} = 5.06\) mm - \(h_{13} = 10.12\) mm - \(h_{14} = 15.18\) mm, etc. > **Summary:** > 1. Use the envelope minima positions to solve for slit width \(a\). > 2. Use the interference maxima formula to get the fringe spacing and calculate the requested distances.