In Young's double slit experiment, the fringe width is 0.4 mm. What is the distance between 4^ th dark band and 6^ th bright band on the same side of the interference pattern?

Step-by-step solution

The fringe width is given as = 0.4 mm. The position of the n^ th bright fringe from the central maximum is given by y_n = n . For the 6^ th bright fringe, y₆ = 6 . The position of the n^ th dark fringe from the central maximum is given by y'_n = (n - 0.5) . For the 4^ th dark fringe, y'₄ = (4 - 0.5) = 3.5 . The distance between the 4^ th dark band and the 6^ th bright band on the same side is: y = y₆ - y'₄ y = 6 - 3.5 = 2.5 Substituting the value of : y = 2.5 0.4 mm = 1.0 mm Answer: 1.0 mm