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#angular width

3 public questions tagged with this topic.

Wavefront and Huygens Principle

What is the angular width of the central maximum in a single-slit diffraction pattern if the slit width is \( 5.0 \, \mu

**Wavefront** is locus of points in same phase, spherical from point source, plane at large distance because radius large, Huygens principle every point on wavefront acts as secondary source of wavelets, new wavefront envelope of secondary wavelets, allows prediction of new wavefront shape from known wavefront, explains reflection and refraction. Angular width 2θ = (2λ/a) . λ = 6.5 × 10⁻⁷ m , a = 5.0 × 10⁻⁶ m . sin θ = (λ/a) = (6.5 × 10⁻⁷/5.0 × 10⁻⁶) = 0.13 , θ = sin⁻¹(0.13) ≈ 7.5° , 2θ ≈ 15° . Using Δ = d sinθ, y = n λ D/d, a sinθ

Ref: NCERT > Physics Book > Wave Optics > Wavefront and Huygens Principle

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Diffraction - Single-Slit and Central Maximum

What is the angular width of the central maximum in a single-slit diffraction pattern if the slit width is \( 12.0 \, \m

**Single-slit pattern** intensity I = I₀ (sinα/α)², α=π a sinθ/λ, central maximum at α=0, minima at α=nπ, so a sinθ=nλ, width increases with λ and D decreases with a, for a=15 μm λ=750 nm first minimum sinθ=750/15000=0.05 θ≈2.87°, angular width of central maximum 2θ≈5.74°. Angular width 2θ = (2λ/a) . λ = 4.8 × 10⁻⁷ m , a = 1.2 × 10⁻⁵ m . sin θ = (λ/a) = (4.8 × 10⁻⁷/1.2 × 10⁻⁵) = 0.04 , θ = sin⁻¹(0.04) ≈ 2.3° , 2θ ≈ 4.6° . Using Δ = d sinθ, y = n λ D/d, a sinθ = n λ, I = I₀

Ref: NCERT > Physics Book > Wave Optics > Diffraction - Single-Slit and Central Maximum

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Diffraction - Single-Slit and Central Maximum

What is the angular width of the central maximum in a single-slit diffraction pattern if the slit width is \( 3.0 \, \mu

**Diffraction bending** property of light waves causes bending around corners, width of central maximum inversely proportional to slit width, intensity of secondary maxima decreases with order because less constructive interference, angular position of minima θ_n = n λ/a, n=±1,±2..., second minimum n=2, third n=3, condition for third secondary maximum approx a sinθ = (2n+1)λ/2. Angular width 2θ = (2λ/a) . λ = 4.5 × 10⁻⁷ m , a = 3.0 × 10⁻⁶ m . sin θ = (λ/a) = (4.5 × 10⁻⁷/3.0 × 10⁻⁶) = 0.15 , θ = sin⁻¹(0.15) ≈ 8.6° , 2θ ≈ 17.2° . Using Δ = d sinθ, y = n

Ref: NCERT > Physics Book > Wave Optics > Diffraction - Single-Slit and Central Maximum

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