What is the angular position of the second minimum in a single-slit diffraction pattern if the slit width is \( 8.0 \, \
**Single-slit diffraction** central maximum width W =2λ D/a, a slit width, D distance, angular width θ =2λ/a, first minimum at a sinθ = λ, fourth minimum a sinθ=4λ, sinθ=4λ/a, for a=5.0 μm λ=500 nm sinθ=4×0.5/5=0.4 θ≈23.6°, central maximum width increases when slit width reduced to half doubles width, when wavelength quadrupled width quadruples, when slit tripled width one-third. Minima occur at sin θ = (nλ/a) . For the second minimum, n = 2 . λ = 6.4 × 10⁻⁷ m , a = 8.0 × 10⁻⁶ m . sin θ = (2 × 6.4 × 10⁻⁷/8.0 × 10⁻⁶) = 0.16 , θ = sin⁻¹(0.16) ≈
Ref: NCERT > Physics Book > Wave Optics > Diffraction - Single-Slit and Central Maximum
