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Superposition, Resultant Amplitude and Intensity

The Superposition, Resultant Amplitude and Intensity category covers the basic principles of wave superposition. It explains how overlapping waves combine, how the resultant amplitude is calculated, and how intensity relates to amplitude. The material is suited for students preparing for physics exams.

30 questions

What is the angular position of the second 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. Secondary maxima occur at θ ≈ ((n + (1/2))λ/a) . For the second secondary maximum, n = 2 . λ = 5.6 × 10⁻⁷ m , a = 7.0 × 10⁻⁶ m . sin θ = ((2 + (1/2)) × 5.6 × 10⁻⁷/7.0 × 10⁻⁶) = (2.5 × 5.6 × 10⁻⁷/7.0 × 10⁻⁶) = 0.2 , θ = sin⁻¹(0.2) ≈ 11.5°

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

In a double-slit experiment, if \( \lambda = 460 \, \text{nm} \), \( d = 0.2 \, \text{mm} \), and \( D = 2.0 \, \text{m}

**Intensity at point** in double-slit I = I_max cos²(φ/2), φ = (2π/λ)Δ, for Δ=λ/4 φ=π/2 I= I_max cos²(π/4)= I_max/2 =2I₀, for Δ=λ/3 φ=2π/3 I= I_max cos²(π/3)= I_max×0.25= I₀, for Δ=5λ/8 φ=5π/4? Actually φ=2π×5/8=5π/4, cos²(5π/8)=?, path difference for destructive φ=(2n+1)π, constructive 2nπ. Bright fringe position x_n = (n λ D/d) . For the third bright fringe, n = 3 . λ = 4.6 × 10⁻⁷ m , d = 2.0 × 10⁻⁴ m , D = 2.0 m . x₃ = (3 × 4.6 × 10⁻⁷ × 2.0/2.0 × 10⁻⁴) = 6.9 × 10⁻³ m = 6.9 mm . Using Δ = d sinθ, y =

Ref: NCERT > Physics Book > Wave Optics > 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

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

What is the intensity of light in the photon picture, according to the wave optics concept?

**Intensity at point** in double-slit I = I_max cos²(φ/2), φ = (2π/λ)Δ, for Δ=λ/4 φ=π/2 I= I_max cos²(π/4)= I_max/2 =2I₀, for Δ=λ/3 φ=2π/3 I= I_max cos²(π/3)= I_max×0.25= I₀, for Δ=5λ/8 φ=5π/4? Actually φ=2π×5/8=5π/4, cos²(5π/8)=?, path difference for destructive φ=(2n+1)π, constructive 2nπ. In the photon picture, intensity is determined by the number of photons crossing a unit area per unit time. 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 Number of photons per unit area per unit time, illustrating interferenc

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

What is the resultant amplitude of two coherent waves of amplitude \( a \) with a phase difference of \( 7\pi/2 \)?

**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). Resultant amplitude A = 2a cos(Φ/2) . For Φ = (7π/2) , A = 2a cos((7π/4)) = 2a ((√(2)/2)) = a√(2) . 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 > Superposition, Resultant Amplitude and Intensity

What is the resultant amplitude when two coherent waves of amplitude \( a \) interfere with a phase difference of \( \pi

**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). Resultant amplitude A = 2a cos(Φ/2) . For Φ = π , A = 2a cos(π/2) = 2a × 0 = 0 . 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 > Superposition, Resultant Amplitude and Intensity

What is the phase difference corresponding to a path difference of \( 2\lambda \) in a double-slit experiment?

**Intensity at point** in double-slit I = I_max cos²(φ/2), φ = (2π/λ)Δ, for Δ=λ/4 φ=π/2 I= I_max cos²(π/4)= I_max/2 =2I₀, for Δ=λ/3 φ=2π/3 I= I_max cos²(π/3)= I_max×0.25= I₀, for Δ=5λ/8 φ=5π/4? Actually φ=2π×5/8=5π/4, cos²(5π/8)=?, path difference for destructive φ=(2n+1)π, constructive 2nπ. Phase difference Φ = (2π/λ) Δ . For Δ = 2λ , Φ = (2π/λ) · 2λ = 4π . 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 4π, illustrating interference, diffraction and polarization principles.

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What is the phase difference corresponding to a path difference of \( 5\lambda/4 \) in a double-slit experiment?

**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. Phase difference Φ = (2π/λ) Δ . For Δ = (5λ/4) , Φ = (2π/λ) · (5λ/4) = (5π/2) . 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 (5π/2), illustrating interference, diffraction and polarization principles.

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

What is the resultant amplitude of two coherent waves of amplitude \( a \) with a phase difference of \( 5\pi/2 \)?

**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). Resultant amplitude A = 2a cos(Φ/2) . For Φ = (5π/2) , A = 2a cos((5π/4)) = 2a (-(√(2)/2)) = -a√(2) , magnitude a√(2) . Using Δ = d sinθ, y = n λ D/d, a sinθ = n λ, I = I₀ cos²θ, n = c/v, sinC = 1/n,

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

What is the intensity at a point in a double-slit experiment where the path difference is \( 5\lambda/2 \), if the maxim

**Intensity at point** in double-slit I = I_max cos²(φ/2), φ = (2π/λ)Δ, for Δ=λ/4 φ=π/2 I= I_max cos²(π/4)= I_max/2 =2I₀, for Δ=λ/3 φ=2π/3 I= I_max cos²(π/3)= I_max×0.25= I₀, for Δ=5λ/8 φ=5π/4? Actually φ=2π×5/8=5π/4, cos²(5π/8)=?, path difference for destructive φ=(2n+1)π, constructive 2nπ. Intensity I = 4I₀ cos²(Φ/2) , where Φ = (2π/λ) Δ . For Δ = (5λ/2) , Φ = (2π/λ) · (5λ/2) = 5π , I = 4I₀ cos²((5π/2)) = 4I₀ × 0 = 0 . 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 > Superposition, Resultant Amplitude and Intensity

What is the intensity at a point in a double-slit experiment where the path difference is \( 7\lambda/2 \), if the maxim

**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). Intensity I = 4I₀ cos²(Φ/2) , where Φ = (2π/λ) Δ . For Δ = (7λ/2) , Φ = (2π/λ) · (7λ/2) = 7π , I = 4I₀ cos²((7π/2)) = 4I₀ × 0 = 0 . Using Δ = d sinθ, y = n λ D/d, a sinθ = n

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