Question

A beam of light consisting of two wavelengths, 4000 Å and 6000 Å, is used to obtain interference fringes in a Young’s double-slit experiment. What is the least distance from the central maximum where the dark fringe is obtained?

Answer 1

Assuming `bb(D/d)` = 10²

λ₁ = 4 × 10⁻⁷ m, λ₂ = 6 × 10⁻⁷ m

Distance at which dark fringe is observed `x = (n + 1/2)(lambdaD)/d`

First Dark fringe for λ₁d₁ = `1/2 (4 xx 10^-7)/(10^-2) m` = 2 × 10⁻⁵ m

First Dark fringe for λ₂d₂ = `1/2(6 xx 10^-7)/(10^-2) m` = 3 × 10⁻⁵ m

The first dark fringe will be the distance where both dark fringes will coincide i.e., the LCM of d₁ and d₁.

i.e. 2 × 10⁻⁵ m × 3 × 10⁻⁵ m

= 6 × 10⁻⁵ m

Answer 2

Question data is insufficient.