single slit diffraction experiment first minimum for λ1 = 660 nm coincides with first maxima for wavelength λ2 . Calculate λ2 .

Q: In a single slit diffraction experiment first minimum for λ₁ = 660 nm coincides with first maxima for wavelength λ₂.Calculate λ₂ .

Sol: position of minima in diffraction pattern is given by; d sin θ = nλ

For first minima of λ₁, we have

d sin θ₁ = 1 × λ₁ ;

or, sin θ₁ = λ₁/d ….(i)

The first maxima approximately lies between first and second minima. For wavelength λ_2 its position will be

d sin θ₂ = (3/2) λ₂

∴ sin θ₂ = (3λ₂)/2d …..(ii)

The two will coincide if,

θ₁ = θ₂

or, sin θ₁ = sin θ₂

∴ λ₁/d = (3λ₂)/2d

λ₂ = (2/3)λ₁ = (2/3) × 660 nm = 440 nm