#intensity calculation

Optical Phenomena and Applications

What is the intensity at a point in a double-slit experiment where the path difference is \( \lambda/3 \), if the maximu

**Incoherent sources** intensity adds I = I₁+I₂, no interference pattern because phase random, two independent sources cannot produce stable interference because phase difference fluctuates rapidly, coherent sources required with constant phase, laser coherent, visibility of fringes requires coherence, degree of coherence determines contrast. Intensity I = 4I₀ cos²(Φ/2) , where Φ = (2π/λ) Δ . For Δ = (λ/3) , Φ = (2π/λ) · (λ/3) = (2π/3) , I = 4I₀ cos²((π/3)) = 4I₀ ((1/2))² = 4I₀ × (1/4) = I₀ . 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 > Optical Phenomena and Applications

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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

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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

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Superposition, Resultant Amplitude and Intensity

What is the intensity at a point in a double-slit experiment where the path difference is \( 3\lambda/4 \), if the maxim

**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. Intensity I = 4I₀ cos²(Φ/2) , where Φ = (2π/λ) Δ . For Δ = (3λ/4) , Φ = (2π/λ) · (3λ/4) = (3π/2) , I = 4I₀ cos²((3π/4)) = 4I₀ ((√(2)/2))² = 4I₀ × (1/2) = 2I₀ . Using Δ = d sinθ, y = n λ D/d, a sinθ = n λ, I = I₀ cos²θ, n = c/v,

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

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Superposition, Resultant Amplitude and Intensity

What is the intensity at a point in a double-slit experiment where the phase difference is \( 2\pi \), if the maximum in

**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) . For Φ = 2π , I = 4I₀ cos²(π) = 4I₀ × 1 = 4I₀ . 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 4I₀, illustrating interference, diffraction and polarization principles.

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

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