In a biprism experiment, a steady interference pattern is observed on the screen kept at a distance of 100 cm using a light of wavelength 5000 Å. Without changing the distance between the virtual images of the slit, the source of light is replaced by a source of wavelength 6400 Å. Now, to reduce the fringe width by 20 % of its initial value, the screen should be moved

Step-by-step solution

The initial fringe width is given by ₁ = ₁ D₁ d The final fringe width is given by ₂ = ₂ D₂ d It is given that the fringe width is reduced by 20 % of its initial value, so: ₂ = ₁ - 0.2 ₁ = 0.8 ₁ Substituting the expressions for fringe widths: ₂ D₂ d = 0.8 ₁ D₁ d D₂ = 0.8 ₁ D₁ ₂ Given ₁ = 5000 , ₂ = 6400 , and D₁ = 100 cm: D₂ = 0.8 5000 100 6400 D₂ = 400000 6400 = 62.5 cm The displacement of the screen is D = D₁ - D₂ = 100 - 62.5 = 37.5 cm. Since D₂ Answer: towards the source by 37.5 cm