Question

A beam of light consisting of two wavelengths 400 nm and 600 nm is used to illuminate a single slit of width 1 mm. Find the least distance of the point from the central maximum where the dark fringes due to both wavelengths coincide on the screen placed 1.5 m from the slit.

Answer 1

Given: λ₁ = 400 nm

λ₂ = 600 nm

a = 1 mm = 10⁻³ m

D = 1.5 m

For coincidence of dark fringes:

m₁λ₁ = m₂λ₂

m₁(400) = m₂(600)

`m_1/m_2 = 600/400`

`m_1/m_2 = 3/2`

m₁ = 3, m₂ = 2

For small angles:

y = `(m lambda D)/a`

= `(m_1 lambda_1 D)/a`

= `(3 xx 400 xx 10^-9 xx 1.5)/10^-3`

= `(1800 xx 10^-9)/10^-3`

= 1.8 × 10⁻³ m

= 1.8 mm

∴ The least distance from the central maximum where the dark fringes of both wavelengths coincide is 1.8 mm.