In a Young's double slit experiment, a parallel beam of monochromatic light is used. The slits are separated by a distance d and the screen is placed parallel to the plane of the slits. If the incident beam makes an angle heta = ⁻¹ ( 2d ) with the normal to the plane of the slits, what will be formed at the centre P₀ of the pattern?

Step-by-step solution

The path difference introduced before the light reaches the slits is x₁ = d . Given = ⁻¹ ( 2d ) , we have = 2d . Substituting the value of , we get x₁ = d ( 2d ) = 2 . At the centre of the screen P₀ , the path difference introduced after the light passes through the slits is x₂ = 0 . The total path difference at P₀ is x = x₁ + x₂ = 2 + 0 = 2 . A path difference of 2 corresponds to a phase difference of , which results in destructive interference. Therefore, a dark fringe will be formed at the centre P₀ .