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#slit separation 0.3 mm

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Wave Properties, Frequency and Energy Conservation

What is the distance of the first bright fringe from the central maximum in a double-slit experiment if \( \lambda = 620

**Intensity not depend on speed** when enters denser medium because intensity I ∝ n E₀²? Actually Poynting vector S = E×H, energy density u =½ ε E², for same amplitude E₀ intensity proportional to n, but amplitude changes at interface due to reflection, total energy conserved incident = reflected + transmitted, interference does not destroy energy, it redistributes. Bright fringe position x_n = (n λ D/d) . For the first bright fringe, n = 1 . λ = 6.2 × 10⁻⁷ m , d = 3.0 × 10⁻⁴ m , D = 1.2 m . x₁ = (1 × 6.2 × 10⁻⁷ × 1.2/3.0 ×

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

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Wave Properties, Frequency and Energy Conservation

What is the fringe width in a double-slit experiment if \( \lambda = 660 \, \text{nm} \), \( d = 0.3 \, \text{mm} \), an

**Frequency of light** remains unchanged when refracts from air into water because frequency determined by source, energy E= h f conserved, speed decreases v= c/n, wavelength decreases λ'=v/f= λ/n, energy of wave proportional to amplitude² not speed, intensity I =½ c ε₀ E₀², energy not depend on speed directly, when speed decreases amplitude may change but energy conserved, interference redistributes energy, total energy same, bright regions gain from dark. Fringe width β = (λ D/d) . λ = 6.6 × 10⁻⁷ m , d = 3.0 × 10⁻⁴ m , D = 1.5 m . β = (6.6 × 10⁻⁷ × 1.5/3.0 × 10⁻⁴) =

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

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