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 = $\frac{\lambda D}{d}$

For λ₁ W₁ = $\frac{{\lambda }_{1}D}{d}$ ……. (1) and

For λ₂ W₂ = $\frac{{\lambda }_{2}D}{d}$ ……. (2)

$\frac{W_2}{W-1}=\frac{{\lambda }_{2}D/d}{{\lambda }_{1}D/d}$

∴ W₂ = $\frac{{\lambda }_2}{{\lambda }_1}{W}_1$ = $\frac{4.8\times {10}^{-7}}{6\times {10}^{-7}}(3.2\times {10}^{-4})$

= 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.