Question

A slit of width 'a' is illuminated by light of wavelength 6000A for what value a will be i. first maximum fall at an angle of diffraction 30 ii. first minimum falls at an angle of diffraction 30

Answer 1

first minimum (dark fringe) at  a sin θ = n λ  ,    where n = 1
            so  sin θ = λ / a
     a = 6000 * 10⁻¹⁰ m * 2 = 1.2 μ meter 

first maximum (after central bright region and first dark fringe) falls at
        sinθ = (n+1/2) λ/a ,        where  n = 1
         a = 3/2 * λ / sinθ = 1.8 μm