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#fringe distance

2 public questions tagged with this topic.

In a double-slit experiment, if \( \lambda = 510 \, \text{nm} \), \( d = 0.3 \, \text{mm} \), and \( D = 1.5 \, \text{m}

**Coherence** requires constant phase difference, two independent sources not coherent because phase random, prevents interference pattern, laser coherent, visibility of fringes determined by coherence, intensity and phase difference, coherent sources produce stable interference, incoherent intensity adds I = I₁+I₂ no interference, interference pattern distinguishes from diffraction. Bright fringe position x_n = (n λ D/d) . For the second bright fringe, n = 2 . λ = 5.1 × 10⁻⁷ m , d = 3.0 × 10⁻⁴ m , D = 1.5 m . x₂ = (2 × 5.1 × 10⁻⁷ × 1.5/3.0 × 10⁻⁴) = 5.1 × 10⁻³ m = 5.1 mm . Using Δ

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

In a double-slit experiment, if \( \lambda = 580 \, \text{nm} \), \( d = 0.2 \, \text{mm} \), and \( D = 1.0 \, \text{m}

**Interference** occurs when two coherent waves superpose, path difference Δ = d sinθ, bright fringe when Δ = n λ, n integer, seventh bright Δ=7λ, dark fringe Δ = (2n-1)λ/2, second dark Δ=3λ/2, third dark 5λ/2, sixth dark 11λ/2, distance of bright fringe from central y = n λ D/d, D screen distance, d slit separation, for λ=650 nm d=0.5 mm D=1.0 m fifth dark? Actually dark y=(2n-1)λ D/(2d). Bright fringe position x_n = (n λ D/d) . For the fourth bright fringe, n = 4 . λ = 5.8 × 10⁻⁷ m , d = 2.0 × 10⁻⁴ m , D = 1.0 m

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