Question

The distance between two consecutive bright fringes in a biprism experiment using light of wavelength 6000 Å is 0.32 mm by how much will the distance change if light of wavelength 4800 Å is used?

Answer 1

Given : λ₁ = 6000 Å = 6 x 10⁻⁷ m, λ₂ = 4800 Å = 4.8 x 10⁻⁷ m,

W₁ = 0.32 mm =3.2 x 10⁻⁴ m

Distance between consecutive bright fringes,

W = \(\frac{λD}{d}\)

For λ₁ W₁ = \(\frac{λ_1D}{d}\) ……. (1) and

For λ₂ W₂ = \(\frac{λ_2D}{d}\) ……. (2)

\(\frac{W_2}{W-1} = \frac{λ_2D/d}{λ_1D/d}\)

∴ W₂ = \(\frac{λ_2}{λ_1}W_1\) = \(\frac{4.8×10^{-7}}{6×10^{-7}}(3.2×10^{-4})\)

= 0.8 x 3.2 x 10⁻⁴

= 2.56 x 10⁻⁴

∴ ΔW = W₁ − W₂ = 3.2 x 10⁻⁴ − 2.56 x 10⁻⁴ =0.64 x 10⁻⁴ m

= 0.064 mm

This is the required change in distance.