The central fringe of the interference pattern produced by light of wavelength 6000 A is shifted to the position of the 4th dark fringe by introducing a thin glass plate of refractive index 1.5 in the path of one ray. What will be the thickness of the glass plate?

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(2) 4.2×10⁻⁶ m. Given: λ = 6000 A = 6×10⁻⁷ m, μ = 1.5, and the plate moves the central fringe out to where the 4th dark fringe used to be. Idea: a sheet of thickness t and index μ in one arm replaces a length t of air by a length μ t of glass, adding an extra optical path (μ-1)t to that ray. The whole pattern slides along until the point that already had that path difference becomes the new centre. The 4th dark fringe lies where the path difference is (2n-1)λ/2 with n = 4, that is (7λ)/2 = 3.5λ. So (μ-1)t = 3.5λ, and μ-1 = 0.5. 0.5 t = 3.5×6×10⁻⁷ = 2.1×10⁻⁶ m. t = (2.1×10⁻⁶)/(0.5) = 4.2×10⁻⁶ m. So the glass plate is 4.2×10⁻⁶ m thick, about 4 micrometre.

The central fringe of the interference pattern produced by light of wavelength 6000 A is shifted to the position of the 4th dark fringe by introducing a thin glass plate of refractive index 1.5 in the path of one ray. What will be the thickness of the glass plate?

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