#single-slit

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.

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

View Details

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

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

View Details

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,

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

View Details

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

Ref: NCERT > Physics Book > Wave Optics > Interference - Double-Slit and Coherence

View Details

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

Ref: NCERT > Physics Book > Wave Optics > Diffraction - Single-Slit and Central Maximum

View Details

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.

Ref: NCERT > Physics Book > Wave Optics > Diffraction - Single-Slit and Central Maximum

View Details

Diffraction - Single-Slit and Central Maximum

What is the angular position of the fourth minimum in a single-slit diffraction pattern if the slit width is \( 5.0 \, \

**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 fourth minimum, n = 4 . λ = 5.0 × 10⁻⁷ m , a = 5.0 × 10⁻⁶ m . sin θ = (4 × 5.0 × 10⁻⁷/5.0 × 10⁻⁶) = 0.4 , θ = sin⁻¹(0.4) ≈ 23.6° . Using Δ

Ref: NCERT > Physics Book > Wave Optics > Diffraction - Single-Slit and Central Maximum

View Details

Diffraction - Single-Slit and Central Maximum

What distinguishes the interference pattern of a double-slit experiment from the diffraction pattern of a single slit?

**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. Interference involves superposition from two sources, producing evenly spaced fringes, while diffraction from one slit creates a broad central maximum with weaker secondary maxima. Using Δ = d sinθ, y = n λ D/d, a sinθ

Ref: NCERT > Physics Book > Wave Optics > Diffraction - Single-Slit and Central Maximum

View Details