Q: In a single slit diffraction experiment first minimum for λ_{1} = 660 nm coincides with first maxima for wavelength λ_{2}.Calculate λ_{2} .

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

For first minima of λ_{1}, we have

d sin θ_{1} = 1 × λ_{1} ;

or, sin θ_{1} = λ_{1}/d ….(i)

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

d sin θ_{2} = (3/2) λ_{2}

∴ sin θ_{2} = (3λ_{2})/2d …..(ii)

The two will coincide if,

θ_{1} = θ_{2}

or, sin θ_{1} = sin θ_{2}

∴ λ_{1}/d = (3λ_{2})/2d

λ_{2} = (2/3)λ_{1} = (2/3) × 660 nm = 440 nm