Diffraction - Single-Slit and Central Maximum

Diffraction - Single-Slit and Central Maximum

In a double-slit experiment, if \( \lambda = 450 \, \text{nm} \), \( d = 0.15 \, \text{mm} \), and \( D = 1.5 \, \text{m

**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. Fringe width β = (λ D/d) . λ = 4.5 × 10⁻⁷ m , d = 1.5 × 10⁻⁴ m , D = 1.5 m . β = (4.5 × 10⁻⁷ × 1.5/1.5 × 10⁻⁴) = 4.5 × 10⁻³ m = 4.5 mm . Using Δ =

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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

In a double-slit experiment, if the wavelength is tripled, what happens to the fringe width?

**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. Fringe width β = (λ D/d) . If λ is tripled, β triples. 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 Triples, illustrating interference, diffraction and polarization principles.

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Diffraction - Single-Slit and Central Maximum

In a single-slit diffraction pattern, what happens to the intensity of the central maximum if the slit width is doubled?

**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. Intensity of the central maximum is proportional to a² . If a is doubled, intensity increases by a factor of 4. Using Δ = d sinθ, y = n λ D/d, a sinθ = n λ, I = I₀ cos²θ, n = c/v, sinC = 1/n, λ'

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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

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Diffraction - Single-Slit and Central Maximum

In a diffraction experiment, if the slit width is \( 4.0 \, \mu\text{m} \) and the wavelength is \( 800 \, \text{nm} \),

**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. First minimum occurs at sin θ = (λ/a) . λ = 800 nm = 8.0 × 10⁻⁷ m , a = 4.0 μm = 4.0 × 10⁻⁶ m . sin θ = (8.0 × 10⁻⁷/4.0 × 10⁻⁶) = 0.2 , so θ = sin⁻¹(0.2) ≈ 11.5° . Using Δ = d

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Diffraction - Single-Slit and Central Maximum

What causes the secondary maxima in a single-slit diffraction pattern to be weaker than the central maximum?

**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 result from partial constructive interference of secondary wavelets, with more cancellations than the fully in-phase central maximum, reducing intensity. 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 Partial interference of wavelets, illustrating interference, diffraction and polarization principles.

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Diffraction - Single-Slit and Central Maximum

What is the condition for the central 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 central maximum occurs at θ = 0° , where the path difference is zero and intensity is maximum. 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

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Diffraction - Single-Slit and Central Maximum

Why does the wave theory predict a change in wavelength but not frequency during refraction?

**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. Frequency is source-dependent and constant, while wavelength adjusts inversely to the speed change in the medium (v = fλ), maintaining the wave equation. Using Δ = d sinθ, y = n λ D/d, a sinθ = n λ, I = I₀ cos²θ, n = c/v, sinC =

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Diffraction - Single-Slit and Central Maximum

What happens to the central maximum’s width in a single-slit diffraction pattern if the wavelength is doubled?

**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) . If λ doubles, the width doubles. 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 Doubles, illustrating interference, diffraction and polarization principles.

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Diffraction - Single-Slit and Central Maximum

What happens to the fringe width in a double-slit experiment if the wavelength of light is halved?

**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. Fringe width β = (λ D/d) . If λ is halved, β is halved. 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 Halves, illustrating interference, diffraction and polarization principles.

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