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#central maximum

13 public questions tagged with this topic.

Wave Properties, Frequency and Energy Conservation

In a diffraction pattern, what happens to the angular width of the central maximum if the slit width is doubled?

**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. Angular width of the central maximum is 2θ = (2λ/a) . If a is doubled, 2θ is halved. 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 Halves, illustrating interference, diffraction and polarization principles.

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

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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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Interference - Double-Slit and Coherence

What explains the presence of a central bright fringe in a single-slit diffraction pattern?

**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. All secondary wavelets from the slit interfere constructively at the center (zero angle), producing a bright fringe due to no path difference. 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), calculatio

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

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Diffraction - Single-Slit and Central Maximum

In a single-slit diffraction pattern, what happens to the intensity of the central maximum if the slit width is doubled?

**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. Intensity of the central maximum is proportional to a² . If a is doubled, intensity increases by a factor of 4. 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 > Diffraction - Single-Slit and Central Maximum

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Diffraction - Single-Slit and Central Maximum

What is the angular width of the central maximum in a single-slit diffraction pattern if the slit width is \( 12.0 \, \m

**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°. Angular width 2θ = (2λ/a) . λ = 4.8 × 10⁻⁷ m , a = 1.2 × 10⁻⁵ m . sin θ = (λ/a) = (4.8 × 10⁻⁷/1.2 × 10⁻⁵) = 0.04 , θ = sin⁻¹(0.04) ≈ 2.3° , 2θ ≈ 4.6° . Using Δ = d sinθ, y = n λ D/d, a sinθ = n λ, I = I₀

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

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Diffraction - Single-Slit and Central Maximum

What is the condition for the central 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 central maximum occurs at θ = 0° , where the path difference is zero and intensity is 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

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

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Diffraction - Single-Slit and Central Maximum

What happens to the central maximum’s width in a single-slit diffraction pattern if the wavelength is doubled?

**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°. Angular width 2θ = (2λ/a) . If λ doubles, the width doubles. 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 Doubles, illustrating interference, diffraction and polarization principles.

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

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Interference - Double-Slit and Coherence

In a double-slit experiment, if the intensity at the central maximum is \( 4I_0 \), what is the intensity where the path

**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. Intensity I = 4I₀ cos²(Φ/2) , where Φ = (2π/λ) Δ . For Δ = λ/4 , Φ = (2π/λ) · (λ/4) = (π/2) , I = 4I₀ cos²(π/4) = 4I₀ × (1/2) = 2I₀ . Using Δ = d sinθ, y = n λ D/d, a sinθ = n λ, I = I₀ cos²θ, n = c/v, sinC =

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

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Diffraction - Single-Slit and Central Maximum

What is the angular width of the central maximum in a single-slit diffraction pattern if the slit width is \( 3.0 \, \mu

**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. Angular width 2θ = (2λ/a) . λ = 4.5 × 10⁻⁷ m , a = 3.0 × 10⁻⁶ m . sin θ = (λ/a) = (4.5 × 10⁻⁷/3.0 × 10⁻⁶) = 0.15 , θ = sin⁻¹(0.15) ≈ 8.6° , 2θ ≈ 17.2° . Using Δ = d sinθ, y = n

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Diffraction - Single-Slit and Central Maximum

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

**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. Angular width 2θ = (2λ/a) . If λ is quadrupled, 2θ increases four times. 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 > 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 central maximum’s width if the slit width is tripled?

**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. Angular width 2θ = (2λ/a) . If a is tripled, 2θ is reduced 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 gives Reduces

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