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#secondary maxima

3 public questions tagged with this topic.

Diffraction - Single-Slit and Central Maximum

What causes the secondary maxima in a single-slit diffraction pattern to be weaker than the central maximum?

**Single-slit pattern** intensity I = I₀ (sinα/α)², α=π a sinθ/λ, central maximum at α=0, minima at α=nπ, so a sinθ=nλ, width increases with λ and D decreases with a, for a=15 μm λ=750 nm first minimum sinθ=750/15000=0.05 θ≈2.87°, angular width of central maximum 2θ≈5.74°. Secondary maxima result from partial constructive interference of secondary wavelets, with more cancellations than the fully in-phase central maximum, reducing intensity. 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 Partial interference of wav

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

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Diffraction - Single-Slit and Central Maximum

Why does the diffraction pattern of a single slit show a central maximum broader than its secondary maxima?

**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. The central maximum results from constructive interference of all secondary wavelets in phase, while secondary maxima involve partial cancellations, reducing their width and intensity. Using Δ = d sinθ, y = n λ D/d, a s

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

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Diffraction - Single-Slit and Central Maximum

In a single-slit diffraction pattern, what happens to the intensity of secondary maxima as their order increases?

**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. The intensity of secondary maxima decreases with increasing order, becoming weaker away from the central maximum. 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

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

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