#diffraction

Optical Phenomena and Applications

Why does the wave nature of light allow it to produce a pattern of bright and dark regions when passing through a narrow

**Incoherent sources** intensity adds I = I₁+I₂, no interference pattern because phase random, two independent sources cannot produce stable interference because phase difference fluctuates rapidly, coherent sources required with constant phase, laser coherent, visibility of fringes requires coherence, degree of coherence determines contrast. Light bends and spreads as waves, with secondary wavelets interfering constructively and destructively, creating diffraction patterns of bright and dark regions. 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

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Wave Properties, Frequency and Energy Conservation

What causes the intensity of light to be zero at certain points in a diffraction pattern?

**Wave model predicts** light bends away from normal when entering rarer medium because speed increases, Snell's law n₁ sinθ₁ = n₂ sinθ₂, n₁>n₂ so sinθ₂>sinθ₁ θ₂>θ₁ away from normal, towards normal when denser, wavefront slows in denser, Huygens construction shows bending. Complete destructive interference occurs when secondary wavelets from different parts of the slit cancel each other out, resulting in zero intensity at minima. 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 Destructive interference, illustrating

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Wave Properties, Frequency and Energy Conservation

What is the condition for the fifth minimum in a single-slit diffraction pattern?

**Wave model predicts** light bends away from normal when entering rarer medium because speed increases, Snell's law n₁ sinθ₁ = n₂ sinθ₂, n₁>n₂ so sinθ₂>sinθ₁ θ₂>θ₁ away from normal, towards normal when denser, wavefront slows in denser, Huygens construction shows bending. Minima occur at sin θ = (nλ/a) . For the fifth minimum, n = 5 , so θ = sin⁻¹((5λ/a)) . 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/a), illustrating interference, diffraction and polarization principles.

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Wave Properties, Frequency and Energy Conservation

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

**Wave model predicts** light bends away from normal when entering rarer medium because speed increases, Snell's law n₁ sinθ₁ = n₂ sinθ₂, n₁>n₂ so sinθ₂>sinθ₁ θ₂>θ₁ away from normal, towards normal when denser, wavefront slows in denser, Huygens construction shows bending. Angular width of the central maximum is 2θ = (2λ/a) . If a is doubled, 2θ 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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Optical Phenomena and Applications

What property of light waves is responsible for the redistribution of energy into bright and dark regions during diffrac

**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. The superposition of secondary wavelets causes constructive and destructive interference, redistributing energy into bright and dark regions. 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), calculat

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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

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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 \( 4.0 \, \m

**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 = 4.0 × 10⁻⁷ m , a = 4.0 × 10⁻⁶ m . sin θ = (3 × 4.0 × 10⁻⁷/4.0 × 10⁻⁶) = 0.3 , θ = sin⁻¹(0.3) ≈ 17.5° . Using Δ = d sinθ, y = n λ D/d, a sinθ = n λ,

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Wavefront and Huygens Principle

Why does the interference pattern from two slits vanish if one slit is covered?

**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. Interference requires superposition from two sources; covering one slit eliminates the second wave, leaving only a diffraction pattern from the single slit. 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 On

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Interference - Double-Slit and Coherence

What explains the presence of a central bright fringe in a single-slit diffraction pattern?

**Double-slit vs single-slit** double-slit interference pattern has equally spaced bright fringes with envelope due to single-slit diffraction, single-slit central maximum width 2λ D/a, intensity of secondary maxima decreases with order, condition for coherence constant frequency and phase, path difference for bright n λ, dark (n+½)λ. All secondary wavelets from the slit interfere constructively at the center (zero angle), producing a bright fringe due to no path difference. 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), calculatio

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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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Wavefront and Huygens Principle

What explains the presence of alternate bright and dark bands near the edge of a shadow in diffraction?

**Huygens principle** predicts shape of wavefront after propagation, for point source close spherical, far plane, after reflection from plane mirror spherical wave becomes spherical with centre mirrored, plane wave remains plane but direction changes angle of incidence equals reflection, after passing through thin prism plane wavefront tilts due to different path. Diffraction causes light to bend around edges, with secondary wavelets interfering constructively and destructively, forming bright and dark bands. 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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