Polarization and Malus Law

Polarization and Malus Law

What is the fringe width in a double-slit experiment if \( \lambda = 590 \, \text{nm} \), \( d = 0.25 \, \text{mm} \), a

**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. Fringe width β = (λ D/d) . λ = 5.9 × 10⁻⁷ m , d = 2.5 × 10⁻⁴ m , D = 1.2 m . β = (5.9 × 10⁻⁷ × 1.2/2.5 × 10⁻⁴) = 2.832 × 10⁻³ m = 2.83 mm . Using Δ = d sinθ, y = n

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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 is the direction of energy propagation relative to the wavefront in a light wave?

**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°. The energy of a wave travels perpendicular to the wavefront. Using Δ = d sinθ, y = n λ D/d, a sinθ = n λ, I = I₀ cos²θ, n = c/v, sinC =

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Polarization and Malus Law

What happens to the fringe width in a double-slit experiment if the slit separation is halved?

**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. Fringe width β = (λ D/d) . If d is halved, β 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 =

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Polarization and Malus Law

Why does the wave theory struggle to explain the propagation of light through a vacuum?

**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. Traditionally, waves require a medium to propagate, and the absence of a medium in a vacuum posed a challenge until the electromagnetic theory showed light as self-sustaining electric and magnetic fields. 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 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 out. Using Δ = d sinθ, y = n λ D/d, a sinθ = n

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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, I = (I₀/4)

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Polarization and Malus Law

What is the direction of the electric field in a linearly polarized light wave relative to its propagation direction?

**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. In a linearly polarized light wave, the electric field oscillates perpendicular to the direction of propagation. 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 Perpendicular,

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Polarization and Malus Law

What is the intensity of light after passing through two polaroids with pass-axes at \( 75^\circ \), 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 75° , I = (I₀/2) cos² 75° . cos 75° ≈ 0.259 , I = (I₀/2) × (0.259)² = (I₀/2) × 0.067

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Polarization and Malus Law

What allows light to continue propagating as a transverse wave after passing through a polaroid?

**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 transverse nature of light, with electric fields oscillating perpendicular to propagation, is preserved, but restricted to the polaroid’s pass-axis direction. 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 =

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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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