Young's double slit experiment manifests the
Young's experiment manifests wave nature of light via interference pattern.
Ref: NCERT Physics Class 12 Chapter 10 Wave Optics
#double slit
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In a double-slit experiment, if the screen is moved closer to the slits, what happens to the fringe width?
**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. Fringe width β = (λ D/d) . If D decreases, β decreases. 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 Decreases, illustrating inter
Ref: NCERT > Physics Book > Wave Optics > Interference - Double-Slit and Coherence
What is the distance of the first dark fringe from the central maximum in a double-slit experiment if \( \lambda = 540 \
**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). Dark fringe position x_n = ((n + (1/2)) λ D/d) . For the first dark fringe, n = 0 . λ = 5.4 × 10⁻⁷ m , d = 2.0 × 10⁻⁴ m , D =
Ref: NCERT > Physics Book > Wave Optics > Interference - Double-Slit and Coherence
What is the distance of the second bright fringe from the central maximum in a double-slit experiment if \( \lambda = 67
**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 = 6.7 × 10⁻⁷ m , d = 6.0 × 10⁻⁴ m , D = 1.8 m . x₂ = (2 × 6.7 × 10⁻⁷ × 1.8/6.0 × 10⁻⁴) = 4.02 × 10⁻³ m = 4.02 mm . Using Δ
Ref: NCERT > Physics Book > Wave Optics > Interference - Double-Slit and Coherence
What happens to the fringe width in a double-slit experiment if the distance between the slits and the screen is tripled
**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). Fringe width β = (λ D/d) . If D is tripled, β triples. Using Δ = d sinθ, y = n λ D/d, a sinθ = n λ, I = I₀ cos²θ, n = c/v, sinC
Ref: NCERT > Physics Book > Wave Optics > Interference - Double-Slit and Coherence
What is the distance of the fifth dark fringe from the central maximum in a double-slit experiment if \( \lambda = 650 \
**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. Dark fringe position x_n = ((n + (1/2)) λ D/d) . For the fifth dark fringe, n = 4 . λ = 6.5 × 10⁻⁷ m , d = 5.0 × 10⁻⁴ m , D = 1.0 m . x₄ = ((4 + (1/2)) × 6.5 × 10⁻⁷ × 1.0/5.0 × 10⁻⁴) = (4.5 × 6.5 × 10⁻⁷/5.0 × 10⁻⁴)
Ref: NCERT > Physics Book > Wave Optics > Interference - Double-Slit and Coherence
What is the path difference for the seventh bright fringe in a double-slit experiment?
**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. Constructive interference occurs at Δ = nλ . For the seventh bright fringe, n = 7 , so Δ = 7λ . 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 7λ, illustrating interference, diffract
Ref: NCERT > Physics Book > Wave Optics > Interference - Double-Slit and Coherence
What determines the visibility of interference fringes when light passes through two slits?
**Superposition principle** resultant displacement sum of individual, for two coherent waves amplitude a each, resultant amplitude A = √(a² + a² +2a² cosφ)=2a|cos(φ/2)|, phase difference φ, path difference Δ = (φ/2π)λ, for φ=π/2 A=√2 a, for φ=6π cos3π=-1? Actually φ=6π cos3π? A=2a|cos3π|=2a, for φ=4π A=2a, intensity I ∝ A², maximum I_max=4I₀ when φ=0, I=2I₀(1+cosφ)=4I₀ cos²(φ/2). Coherence, or a constant phase difference between the waves from the slits, is essential for a visible, stable interference pattern. Using Δ = d sinθ, y = n λ D/d, a sinθ = n λ, I = I₀ cos²θ, n = c/v, sinC = 1/n, λ' =
Ref: NCERT > Physics Book > Wave Optics > Superposition, Resultant Amplitude and Intensity
