#diffraction pattern

Optical Phenomena and Applications

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

**Convex lens focusing** plane wave into point because lens introduces phase delay proportional to thickness, converting plane wavefront to spherical converging to focal point, property ensures rays parallel to axis meet at focus, spherical aberration minimized for paraxial rays, lensmaker's formula determines focal length. Secondary maxima occur at θ ≈ ((n + (1/2))λ/a) . For the third secondary maximum, n = 3 , θ ≈ (7λ/2a) . 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 θ = (7λ/2a),

Ref: NCERT > Physics Book > Wave Optics > Optical Phenomena and Applications

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Superposition, Resultant Amplitude and Intensity

In a single-slit diffraction pattern, what happens to the central maximum’s width if the wavelength is reduced to one-th

**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). Angular width 2θ = (2λ/a) . If λ is reduced to one-third, 2θ reduces to one-third. 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

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 fourth secondary maximum in a single-slit diffraction pattern?

**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 occur at θ ≈ ((n + (1/2))λ/a) . For the fourth secondary maximum, n = 4 , θ ≈ (9λ/2a) . 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 θ = (9λ/2a),

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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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