#light intensity

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

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Refraction, Refractive Index and Critical Angle

What determines the intensity of light in the wave picture?

**Refractive index** n = c/v, c=3×10⁸ m/s vacuum, v speed in medium, for v=2.25×10⁸ m/s n=3/2.25=1.33, for v=1.8×10⁸ n=1.67, for n=1.3 v=3×10⁸/1.3=2.31×10⁸ m/s, for n=1.6 v=1.875×10⁸ m/s, for n=1.2 λ_medium = λ_vacuum/n =600/1.2=500 nm, for n=1.75 λ=700/1.75=400 nm, for n=1.5 λ=750/1.5=500 nm, wavelength in medium λ' = λ/n. In the wave picture, intensity is proportional to the square of the amplitude of the wave. 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 Square of the amplitude, illustrating interference, dif

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Polarization and Malus Law

Why does the intensity of light transmitted through two polaroids become zero when their axes are at 90° to each other?

**Polarization** transverse wave property, light after polaroid polarized along pass-axis, Malus law I = I₀ cos²θ, θ angle between pass-axes, initial unpolarized intensity I₀ after first polaroid I₁ = I₀/2, after second at 45° I₂ = I₁ cos²45°= I₀/2×0.5= I₀/4, after two perpendicular 90° I=0 because cos90°=0, for 60° I= I₀/2×cos²60°= I₀/2×0.25= I₀/8. The first polaroid aligns the electric field, and the second, perpendicular to it, blocks all components, as the cosine of 90° is zero in the intensity relation. Using Δ = d sinθ, y = n λ D/d, a sinθ = n λ, I = I₀ cos²θ, n = c/v, sinC = 1/n,

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Polarization and Malus Law

What causes the intensity of light to remain conserved during interference despite the presence of dark fringes?

**Two polaroids** with pass-axes perpendicular intensity zero because second blocks component, with 45° intensity I₀/4 if initial after first I₀, if initial unpolarized I₀, after first I₀/2 then after second at 45° I₀/4. Three polaroids first and third crossed 90°, middle at 45° maximum transmission because middle rotates polarization, I after first I₀/2, after middle at 45° I₀/4, after third at 45° to middle I₀/4×cos²45°= I₀/8, non-zero, maximum when middle 45°. Energy is redistributed from dark to bright fringes through interference, conserving total energy as the sum of intensities balances

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Polarization and Malus Law

What is the intensity of light after passing through three polaroids, with the first and third crossed and the second at

**Two polaroids** with pass-axes perpendicular intensity zero because second blocks component, with 45° intensity I₀/4 if initial after first I₀, if initial unpolarized I₀, after first I₀/2 then after second at 45° I₀/4. Three polaroids first and third crossed 90°, middle at 45° maximum transmission because middle rotates polarization, I after first I₀/2, after middle at 45° I₀/4, after third at 45° to middle I₀/4×cos²45°= I₀/8, non-zero, maximum when middle 45°. After the first polaroid, I = (I₀/2) . After the second at 45° , I = (I₀/2) cos² 45° = (I₀/4) . Third at 90° - 45° = 45° to second,

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Polarization and Malus Law

What is the intensity of light after passing through a polaroid rotated at \( 30^\circ \) relative to the initial polari

**Polarization** transverse wave property, light after polaroid polarized along pass-axis, Malus law I = I₀ cos²θ, θ angle between pass-axes, initial unpolarized intensity I₀ after first polaroid I₁ = I₀/2, after second at 45° I₂ = I₁ cos²45°= I₀/2×0.5= I₀/4, after two perpendicular 90° I=0 because cos90°=0, for 60° I= I₀/2×cos²60°= I₀/2×0.25= I₀/8. Using Malus’ law, I = I₀ cos² θ . For θ = 30° , cos 30° = (√(3)/2) , I = I₀ ((√(3)/2))² = I₀ × (3/4) = 0.75 I₀ . Using Δ = d sinθ, y = n λ D/d, a sinθ = n λ, I = I₀ cos²θ, n =

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Polarization and Malus Law

What is the intensity of light after passing through two polaroids with pass-axes at \( 60^\circ \), if the intensity af

**Polaroid rotation** intensity varies sinusoidally with angle due to Malus law I = I₀ cos²θ, when polaroid rotated 90° from initial, intensity goes from max to zero, for unpolarized light rotating polaroid does not change intensity after first polaroid because average, but second polaroid intensity depends on relative angle, explains why intensity transmitted through two polaroids drops to zero when perpendicular. Using Malus’ law, I = I₀ cos² θ . For θ = 60° , cos 60° = 0.5 , I = I₀ (0.5)² = 0.25 I₀ . Using Δ = d sinθ, y = n λ D/d, a sinθ = n λ, I =

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Polarization and Malus Law

What is the intensity of light after passing through two polaroids with pass-axes at \( 15^\circ \), if the initial unpo

**Two polaroids** with pass-axes perpendicular intensity zero because second blocks component, with 45° intensity I₀/4 if initial after first I₀, if initial unpolarized I₀, after first I₀/2 then after second at 45° I₀/4. Three polaroids first and third crossed 90°, middle at 45° maximum transmission because middle rotates polarization, I after first I₀/2, after middle at 45° I₀/4, after third at 45° to middle I₀/4×cos²45°= I₀/8, non-zero, maximum when middle 45°. After the first polaroid, I = (I₀/2) . After the second at 15° , I = (I₀/2) cos² 15° . cos 15° ≈ 0.966 , I = (I₀/2) × (0.966)² = (I₀

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Polarization and Malus Law

Why does the intensity of light remain unaffected by changes in its speed during refraction?

**Polaroid rotation** intensity varies sinusoidally with angle due to Malus law I = I₀ cos²θ, when polaroid rotated 90° from initial, intensity goes from max to zero, for unpolarized light rotating polaroid does not change intensity after first polaroid because average, but second polaroid intensity depends on relative angle, explains why intensity transmitted through two polaroids drops to zero when perpendicular. Intensity depends on the square of the amplitude, which remains constant during refraction, while speed changes do not alter the energy per unit area. Using Δ = d sinθ, y = n λ D/d,

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Polarization and Malus Law

Why does the intensity of light remain unchanged in terms of energy when it undergoes interference or diffraction?

**Two polaroids** with pass-axes perpendicular intensity zero because second blocks component, with 45° intensity I₀/4 if initial after first I₀, if initial unpolarized I₀, after first I₀/2 then after second at 45° I₀/4. Three polaroids first and third crossed 90°, middle at 45° maximum transmission because middle rotates polarization, I after first I₀/2, after middle at 45° I₀/4, after third at 45° to middle I₀/4×cos²45°= I₀/8, non-zero, maximum when middle 45°. Interference and diffraction redistribute light energy without loss, as bright and dark regions balance out, conserving total energy

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Polarization and Malus Law

What happens to the intensity of light when it passes through a polaroid and the polaroid is rotated by \( 90^\circ \) f

**Polarization** transverse wave property, light after polaroid polarized along pass-axis, Malus law I = I₀ cos²θ, θ angle between pass-axes, initial unpolarized intensity I₀ after first polaroid I₁ = I₀/2, after second at 45° I₂ = I₁ cos²45°= I₀/2×0.5= I₀/4, after two perpendicular 90° I=0 because cos90°=0, for 60° I= I₀/2×cos²60°= I₀/2×0.25= I₀/8. For unpolarized light, intensity after one polaroid is (I₀/2) . Rotating by 90° from the pass-axis gives I = (I₀/2) cos² 90° = 0 . It reduces to zero from its initial value. Using Δ = d sinθ, y = n λ D/d, a sinθ = n λ, I = I₀ cos²θ,

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Polarization and Malus Law

Why does the intensity of light transmitted through two polaroids drop to zero when their pass-axes are perpendicular?

**Polarization** transverse wave property, light after polaroid polarized along pass-axis, Malus law I = I₀ cos²θ, θ angle between pass-axes, initial unpolarized intensity I₀ after first polaroid I₁ = I₀/2, after second at 45° I₂ = I₁ cos²45°= I₀/2×0.5= I₀/4, after two perpendicular 90° I=0 because cos90°=0, for 60° I= I₀/2×cos²60°= I₀/2×0.25= I₀/8. When pass-axes are perpendicular, the electric field component along the second polaroid’s axis is zero (cos 90° = 0), blocking all light per Malus’ law. Using Δ = d sinθ, y = n λ D/d, a sinθ = n λ, I = I₀ cos²θ, n = c/v, sinC = 1/n, λ' =

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