Light of wavelength 630 nm passes through a grating and then onto a screen located several meters from the grating. The 1st order bright band is located 0.26m from the central maximum. Light from a second source produces a band 0.21m from the central maximum.

Question

Determine the wavelength of the second source. [Hint: If the angular deflection is small, tan θ = sin θ.]

Answer 1

d•sinφ=kλ,
tanφ=x/L. =>
d•x/L=kλ
λ₁=dx₁/Lk
λ₂=dx₂/Lk
λ₁/λ₂=dx₁Lk/dx₂Lk= x₁/x₂
λ₂=λ₁x₂/x₁=630•0.21/0.26=508 nm