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#wavelength 600 nm

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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

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Wave Properties, Frequency and Energy Conservation

What is the frequency of light with a wavelength of \( 600 \, \text{nm} \) in air, given the speed of light in air is \(

**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. Frequency nu = (c/λ) . λ = 600 nm = 6.0 × 10⁻⁷ m , c = 3.0 × 10⁸ m/s . nu = (3.0 × 10⁸/6.0 × 10⁻⁷) = 5.0 × 10¹⁴ Hz . 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

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

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Superposition, Resultant Amplitude and Intensity

What is the angular position of the first secondary maximum in a single-slit diffraction pattern if the slit width is \(

**Phase difference** corresponding to path difference Δ, φ =2π Δ/λ, for Δ=5λ/8 φ=5π/4, for Δ=9λ/4 φ=9π/2, for Δ=λ path difference φ=2π constructive, but for destructive condition path difference λ can be destructive if one reflection introduces π phase shift, resultant amplitude zero when φ=(2n+1)π. First secondary maximum occurs at θ ≈ (3λ/2a) . λ = 6.0 × 10⁻⁷ m , a = 6.0 × 10⁻⁶ m . sin θ = (3 × 6.0 × 10⁻⁷/2 × 6.0 × 10⁻⁶) = 0.15 , θ = sin⁻¹(0.15) ≈ 8.6° . Using Δ = d sinθ, y = n λ D/d, a sinθ = n λ, I =

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

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