#interference pattern

Wavefront and Huygens Principle

What property of light waves enables the formation of a stable interference pattern when split from a single source?

**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. Coherence, or a fixed phase relationship, is maintained when light from one source is split, allowing constructive and destructive interference to form a stable 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 > Wavefront and Huygens Principle

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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

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Diffraction - Single-Slit and Central Maximum

In a double-slit experiment, if the wavelength is tripled, what happens to the fringe width?

**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. Fringe width β = (λ D/d) . If λ is tripled, β triples. 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 Triples, illustrating interference,

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

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Superposition, Resultant Amplitude and Intensity

Why does the interference pattern disappear if the phase difference between two sources changes rapidly over time?

**Intensity at point** in double-slit I = I_max cos²(φ/2), φ = (2π/λ)Δ, for Δ=λ/4 φ=π/2 I= I_max cos²(π/4)= I_max/2 =2I₀, for Δ=λ/3 φ=2π/3 I= I_max cos²(π/3)= I_max×0.25= I₀, for Δ=5λ/8 φ=5π/4? Actually φ=2π×5/8=5π/4, cos²(5π/8)=?, path difference for destructive φ=(2n+1)π, constructive 2nπ. Rapid phase changes make the sources incoherent, averaging out the interference effects, resulting in a uniform intensity distribution. 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 Sources become incoherent, illustrating int

Ref: NCERT > Physics Book > Wave Optics > Superposition, Resultant Amplitude and Intensity

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Diffraction - Single-Slit and Central Maximum

What is the condition for the first secondary maximum in a single-slit diffraction pattern?

**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. The first secondary maximum occurs approximately at θ = (3λ/2a) , where a is the slit width and λ is the wavelength. Using Δ = d sinθ, y = n λ D/d, a sinθ = n λ, I = I₀ cos²θ, n = c/v, sinC = 1/n,

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

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Interference - Double-Slit and Coherence

What happens to the fringe width in a double-slit experiment if both the slit separation and screen distance are doubled

**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 and d are both doubled, β = (λ (2D)/2d) = (λ D/d) , so it remains the same. 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

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

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Interference - Double-Slit and Coherence

What happens to the fringe width in a double-slit experiment if the slit separation is reduced to one-third?

**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 reduced to (d/3) , β increases to 3β , i.e., triples. Using Δ = d sinθ, y = n λ D/d, a sinθ = n λ,

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

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Interference - Double-Slit and Coherence

What is the path difference for the second dark 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. Destructive interference occurs at Δ = (n + (1/2))λ . For the second dark fringe, n = 1 , Δ = (1 + (1/2))λ = (3λ/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),

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

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