#dark fringe

What is the path difference for the sixth dark fringe in a double-slit experiment?

**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. Destructive interference occurs at Δ = (n + (1/2))λ . For the sixth dark fringe, n = 5 , Δ = (5 + (1/2))λ = (11λ/2) . 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

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

View Details

Optical Phenomena and Applications

What is the path difference for the third dark fringe in a double-slit experiment?

**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. Destructive interference occurs at Δ = (n + (1/2))λ . For the third dark fringe, n = 2 , Δ = (2 + (1/2))λ = (5λ/2) . 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

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

View Details

Optical Phenomena and Applications

What is the distance of the third dark fringe from the central maximum in a double-slit experiment if \( \lambda = 600 \

**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. Dark fringe position x_n = ((n + (1/2)) λ D/d) . For the third dark fringe, n = 2 . λ = 6.0 × 10⁻⁷ m , d = 4.0 × 10⁻⁴ m , D = 2.0 m . x₂ = ((2 + (1/2)) × 6.0 × 10⁻⁷ × 2.0/4.0 × 10⁻⁴) = (2.5 × 1.2 × 10⁻⁶/4.0 × 10⁻⁴) = 7.5

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

View Details

Wave Properties, Frequency and Energy Conservation

What is the path difference for the fifth dark fringe in a double-slit experiment?

**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. Destructive interference occurs at Δ = (n + (1/2))λ . For the fifth dark fringe, n = 4 , Δ = (4 + (1/2))λ = (9λ/2) . 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 (9λ/2), illustrating interference, diffraction and polarization principles.

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

View Details

Interference - Double-Slit and Coherence

What is the distance of the second dark fringe from the central maximum in a double-slit experiment if \( \lambda = 700

**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. For dark fringes, x_n = ((n + (1/2)) λ D/d) . Second dark fringe, n = 1 . λ = 7.0 × 10⁻⁷ m , d = 3.5 × 10⁻⁴ m , D = 1.5 m . x₁ = ((1 + (1/2)) × 7.0 × 10⁻⁷ × 1.5/3.5 × 10⁻⁴) = (1.5 × 1.05 × 10⁻⁶/3.5 × 10⁻⁴) = 4.5