E6. For the same conditions described in exercise 5, at what distance from the center of the screen will the fourth dark fringe appear?

To determine the distance from the center of the screen where the **fourth dark fringe** appears, we need to use the formula for the position of dark fringes in a double-slit interference pattern. However, since the specific conditions from **Exercise 5** are not provided, I'll outline the general approach and formula you can use once you have the necessary parameters. ### General Formula for Dark Fringes The position \( y_m \) of the \( m^{th} \) dark fringe in a double-slit interference pattern is given by: \[ y_m = \left( m + \frac{1}{2} \right) \frac{\lambda D}{d} \] Where: - \( y_m \) = distance from the center of the screen to the \( m^{th} \) dark fringe - \( m \) = order of the dark fringe (for the fourth dark fringe, \( m = 3 \) since \( m \) starts from 0) - \( \lambda \) = wavelength of the light used - \( D \) = distance from the double-slit to the screen - \( d \) = separation between the two slits ### Steps to Calculate 1. **Identify the Parameters:** - **Wavelength (\( \lambda \))**: This should be given in Exercise 5. - **Distance to Screen (\( D \))**: Also provided in Exercise 5. - **Slit Separation (\( d \))**: Provided in Exercise 5. 2. **Determine the Order (\( m \)):** - For the fourth dark fringe, \( m = 3 \) (since \( m = 0 \) corresponds to the first dark fringe). 3. **Plug the Values into the Formula:** \[ y_3 = \left( 3 + \frac{1}{2} \right) \frac{\lambda D}{d} = \frac{7}{2} \cdot \frac{\lambda D}{d} \] 4. **Calculate \( y_3 \):** - Substitute the known values of \( \lambda \), \( D \), and \( d \) into the equation to find the distance from the center to the fourth dark fringe. ### Example Calculation *Assuming* the following values (replace these with the actual values from Exercise 5): - Wavelength, \( \lambda = 600 \) nm \( = 600 \times 10^{-9} \) m - Distance to screen, \( D = 2 \) m - Slit separation, \( d = 0.01 \) m \[ y_3 = \frac{7}{2} \cdot \frac{600 \times 10^{-9} \, \text{m} \times 2 \, \text{m}}{0.01 \, \text{m}} = \frac{7}{2} \cdot \frac{1.2 \times 10^{-6} \, \text{m}^2}{0.01 \, \text{m}} = \frac{7}{2} \times 1.2 \times 10^{-4} \, \text{m} = 4.2 \times 10^{-4} \, \text{m} = 0.42 \, \text{mm} \] So, in this example, the fourth dark fringe would appear **0.42 mm** from the center of the screen. ### Conclusion To find the exact distance for your specific problem: 1. Refer back to **Exercise 5** to obtain the values for \( \lambda \), \( D \), and \( d \). 2. Use the formula provided above with \( m = 3 \) for the fourth dark fringe. 3. Perform the calculation to find \( y_3 \). If you provide the specific values from Exercise 5, I can help you carry out the calculation more precisely!