#single-slit diffraction

What is the angular position of the first minimum in a single-slit diffraction pattern if the slit width is \( 15.0 \, \

**Polarization requires transverse waves** because only transverse can have orientation perpendicular to propagation, longitudinal cannot be polarized, wave theory requires light transverse to explain polarization, polaroids transmit only component along pass-axis, unpolarized has random transverse orientations, after polaroid polarized. First minimum occurs at sin θ = (λ/a) . λ = 7.5 × 10⁻⁷ m , a = 1.5 × 10⁻⁵ m . sin θ = (7.5 × 10⁻⁷/1.5 × 10⁻⁵) = 0.05 , θ = sin⁻¹(0.05) ≈ 2.9° . Using Δ = d sinθ, y = n λ D/d, a sinθ = n λ, I = I₀ cos²θ, n = c/v, sinC = 1/n, λ'

Ref: NCERT > Physics Book > Wave Optics > Optical Phenomena and Applications

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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),

Ref: NCERT > Physics Book > Wave Optics > Optical Phenomena and Applications

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Wave Properties, Frequency and Energy Conservation

What is the angular position of the first minimum in a single-slit diffraction pattern if the slit width is \( 8.0 \, \m

**Intensity not depend on speed** when enters denser medium because intensity I ∝ n E₀²? Actually Poynting vector S = E×H, energy density u =½ ε E², for same amplitude E₀ intensity proportional to n, but amplitude changes at interface due to reflection, total energy conserved incident = reflected + transmitted, interference does not destroy energy, it redistributes. First minimum occurs at sin θ = (λ/a) . λ = 6.4 × 10⁻⁷ m , a = 8.0 × 10⁻⁶ m . sin θ = (6.4 × 10⁻⁷/8.0 × 10⁻⁶) = 0.08 , θ = sin⁻¹(0.08) ≈ 4.6° . Using Δ = d sinθ, y =

Ref: NCERT > Physics Book > Wave Optics > Wave Properties, Frequency and Energy Conservation

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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

In a single-slit diffraction pattern, what happens to the central maximum’s width if the wavelength is reduced to one-th

**Superposition principle** resultant displacement sum of individual, for two coherent waves amplitude a each, resultant amplitude A = √(a² + a² +2a² cosφ)=2a|cos(φ/2)|, phase difference φ, path difference Δ = (φ/2π)λ, for φ=π/2 A=√2 a, for φ=6π cos3π=-1? Actually φ=6π cos3π? A=2a|cos3π|=2a, for φ=4π A=2a, intensity I ∝ A², maximum I_max=4I₀ when φ=0, I=2I₀(1+cosφ)=4I₀ cos²(φ/2). Angular width 2θ = (2λ/a) . If λ is reduced to one-third, 2θ reduces to one-third. 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

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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 =

Ref: NCERT > Physics Book > Wave Optics > Superposition, Resultant Amplitude and Intensity

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Wavefront and Huygens Principle

What is the angular position of the third minimum in a single-slit diffraction pattern if the slit width is \( 10.0 \, \

**Wavefront types** point source spherical, distant point source plane, after convex lens plane wave focuses to point because lens adds phase delay proportional to thickness, converging spherical wavefront, after concave mirror plane wave becomes spherical converging to focus, illustrating Huygens construction. Minima occur at sin θ = (nλ/a) . For the third minimum, n = 3 . λ = 6.0 × 10⁻⁷ m , a = 1.0 × 10⁻⁵ m . sin θ = (3 × 6.0 × 10⁻⁷/1.0 × 10⁻⁵) = 0.18 , θ = sin⁻¹(0.18) ≈ 10.4° . Using Δ = d sinθ, y = n λ D/d, a sinθ = n λ,

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

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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°