#slit width

Optical Phenomena and Applications

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'

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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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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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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,

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Wavefront and Huygens Principle

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

**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. First minimum occurs at sin θ = (λ/a) . λ = 5.0 × 10⁻⁷ m , a = 1.0 × 10⁻⁵ m . sin θ = (5.0 × 10⁻⁷/1.0 × 10⁻⁵) = 0.05 , θ = sin⁻¹(0.05) ≈ 2.9° . Using Δ = d sinθ, y = n λ D/d, a sinθ = n

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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

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 is the angular position of the second minimum in a single-slit diffraction pattern if the slit width is \( 2.0 \, \

**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. Minima occur at sin θ = (nλ/a) . For the second minimum, n = 2 . λ = 4.0 × 10⁻⁷ m , a = 2.0 × 10⁻⁶ m . sin θ = (2 × 4.0 × 10⁻⁷/2.0 × 10⁻⁶) = 0.4 , θ = sin⁻¹(0.4) ≈ 23.6° . Using Δ

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Diffraction - Single-Slit and Central Maximum

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

**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. Minima occur at sin θ = (nλ/a) . For the second minimum, n = 2 . λ = 6.4 × 10⁻⁷ m , a = 8.0 × 10⁻⁶ m . sin θ = (2 × 6.4 × 10⁻⁷/8.0 × 10⁻⁶) = 0.16 , θ = sin⁻¹(0.16) ≈

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Diffraction - Single-Slit and Central Maximum

What is the angular position of the third minimum in a single-slit diffraction pattern if the slit width is \( 6.0 \, \m

**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. Minima occur at sin θ = (nλ/a) . For the third minimum, n = 3 . λ = 4.8 × 10⁻⁷ m , a = 6.0 × 10⁻⁶ m . sin θ = (3 × 4.8 × 10⁻⁷/6.0 × 10⁻⁶) = 0.24 , θ = sin⁻¹(0.24) ≈ 13.9° . Using Δ

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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

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Diffraction - Single-Slit and Central Maximum

What is the angular position of the second minimum in a single-slit diffraction pattern if the slit width is \( 3.0 \, \

**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. Minima occur at sin θ = (nλ/a) . For the second minimum, n = 2 . λ = 6.0 × 10⁻⁷ m , a = 3.0 × 10⁻⁶ m . sin θ = (2 × 6.0 × 10⁻⁷/3.0 × 10⁻⁶) = 0.4 , θ = sin⁻¹(0.4) ≈

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