**Problem 6: Interference from Two Slits**Two parallel slits are illuminated by light composed of two wavelengths. One wavelength is \(\lambda_A = 645 \, \text{nm}\). The other wavelength is \(\lambda_B\) and is unknown. On a viewing screen, the light with wavelength \(\lambda_A = 645 \, \text{nm}\) produces its third-order bright fringe at the same place where the light with wavelength \(\lambda_B\) produces its fourth dark fringe. The fringes are counted relative to the central or zeroth-order bright fringe. What is the unknown wavelength?**Explanation:**To determine the unknown wavelength \(\lambda_B\), we must apply the principles of interference for both bright and dark fringes in a double-slit experiment.### Bright Fringe (Constructive Interference):For \(\lambda_A\), the position of the m-th order bright fringe is given by:\[ d \sin \theta = m \lambda_A \]where:- \(d\) is the distance between the two slits,- \(\theta\) is the angle of the fringe,- \(m\) is the order of the fringe (here, \(m = 3\) for the third bright fringe).### Dark Fringe (Destructive Interference):For \(\lambda_B\), the position of the m-th order dark fringe is given by:\[ d \sin \theta = \left( m + \frac{1}{2} \right) \lambda_B \]where:- \(m\) is the order of the fringe (here, \(m = 4\) for the fourth dark fringe).Since both the bright fringe of \(\lambda_A\) and the dark fringe of \(\lambda_B\) occur at the same location on the screen (\(\sin \theta\) is the same for both), we can set the two equations equal:\[ 3 \lambda_A = \left( 4 + \frac{1}{2} \right) \lambda_B \]\[ 3 \times 645 \, \text{nm} = \frac{9}{2} \lambda_B \]\[ 1935 \, \text{nm} = \frac{9}{2} \lambda_B \]\[ \lambda_B = \frac{1935 \, \text{nm} \times

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