#secondary maximum

Optical Phenomena and Applications

What is the condition for the third secondary maximum in a single-slit diffraction pattern?

**Convex lens focusing** plane wave into point because lens introduces phase delay proportional to thickness, converting plane wavefront to spherical converging to focal point, property ensures rays parallel to axis meet at focus, spherical aberration minimized for paraxial rays, lensmaker's formula determines focal length. Secondary maxima occur at θ ≈ ((n + (1/2))λ/a) . For the third secondary maximum, n = 3 , θ ≈ (7λ/2a) . Using Δ = d sinθ, y = n λ D/d, a sinθ = n λ, I = I₀ cos²θ, n = c/v, sinC = 1/n, λ' = λ/n and A = 2a cos(φ/2), calculation gives θ = (7λ/2a),

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Superposition, Resultant Amplitude and Intensity

What is the angular position of the second secondary maximum in a single-slit diffraction pattern if the slit width is \

**Phase difference** corresponding to path difference Δ, φ =2π Δ/λ, for Δ=5λ/8 φ=5π/4, for Δ=9λ/4 φ=9π/2, for Δ=λ path difference φ=2π constructive, but for destructive condition path difference λ can be destructive if one reflection introduces π phase shift, resultant amplitude zero when φ=(2n+1)π. Secondary maxima occur at θ ≈ ((n + (1/2))λ/a) . For the second secondary maximum, n = 2 . λ = 5.6 × 10⁻⁷ m , a = 7.0 × 10⁻⁶ m . sin θ = ((2 + (1/2)) × 5.6 × 10⁻⁷/7.0 × 10⁻⁶) = (2.5 × 5.6 × 10⁻⁷/7.0 × 10⁻⁶) = 0.2 , θ = sin⁻¹(0.2) ≈ 11.5°

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Superposition, Resultant Amplitude and Intensity

What is the angular position of the first secondary maximum in a single-slit diffraction pattern if the slit width is \(

**Phase difference** corresponding to path difference Δ, φ =2π Δ/λ, for Δ=5λ/8 φ=5π/4, for Δ=9λ/4 φ=9π/2, for Δ=λ path difference φ=2π constructive, but for destructive condition path difference λ can be destructive if one reflection introduces π phase shift, resultant amplitude zero when φ=(2n+1)π. First secondary maximum occurs at θ ≈ (3λ/2a) . λ = 6.0 × 10⁻⁷ m , a = 6.0 × 10⁻⁶ m . sin θ = (3 × 6.0 × 10⁻⁷/2 × 6.0 × 10⁻⁶) = 0.15 , θ = sin⁻¹(0.15) ≈ 8.6° . Using Δ = d sinθ, y = n λ D/d, a sinθ = n λ, I =

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Diffraction - Single-Slit and Central Maximum

What is the angular position of the first secondary maximum in a single-slit diffraction pattern if the slit width is \(

**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°. Secondary maxima occur at θ ≈ ((n + (1/2))λ/a) . For the first secondary maximum, n = 1 . λ = 4.5 × 10⁻⁷ m , a = 9.0 × 10⁻⁶ m . sin θ = ((1 + (1/2)) × 4.5 × 10⁻⁷/9.0 × 10⁻⁶) = (1.5 × 4.5 × 10⁻⁷/9.0 × 10⁻⁶) = 0.075 , θ = sin⁻¹(0.075) ≈ 4.3°

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Diffraction - Single-Slit and Central Maximum

What is the condition for the fourth secondary maximum in a single-slit diffraction pattern?

**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°. Secondary maxima occur at θ ≈ ((n + (1/2))λ/a) . For the fourth secondary maximum, n = 4 , θ ≈ (9λ/2a) . Using Δ = d sinθ, y = n λ D/d, a sinθ = n λ, I = I₀ cos²θ, n = c/v, sinC = 1/n, λ' = λ/n and A = 2a cos(φ/2), calculation gives θ = (9λ/2a),

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Diffraction - Single-Slit and Central Maximum

What is the condition for the first secondary maximum in a single-slit diffraction pattern?

**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. The first secondary maximum occurs approximately at θ = (3λ/2a) , where a is the slit width and λ is the wavelength. Using Δ = d sinθ, y = n λ D/d, a sinθ = n λ, I = I₀ cos²θ, n = c/v, sinC = 1/n,

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Diffraction - Single-Slit and Central Maximum

What is the condition for the second secondary maximum in a single-slit diffraction pattern?

**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°. Secondary maxima occur at θ ≈ ((n + (1/2))λ/a) . For the second secondary maximum, n = 2 , θ ≈ (5λ/2a) . Using Δ = d sinθ, y = n λ D/d, a sinθ = n λ, I = I₀ cos²θ, n = c/v, sinC = 1/n, λ' = λ/n and A = 2a cos(φ/2), calculation gives θ = (5λ/2a),

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