#polaroids

Optical Phenomena and Applications

What causes the intensity of light to vary sinusoidally with angle when passing through two polaroids?

**Polarization requires transverse waves** because only transverse can have orientation perpendicular to propagation, longitudinal cannot be polarized, wave theory requires light transverse to explain polarization, polaroids transmit only component along pass-axis, unpolarized has random transverse orientations, after polaroid polarized. The intensity follows Malus’ law, where it varies as the square of the cosine of the angle between the polaroids’ axes, producing a sinusoidal pattern. 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 > Optical Phenomena and Applications

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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,

Ref: NCERT > Physics Book > Wave Optics > Polarization and Malus Law

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Polarization and Malus Law

What is the intensity of light after passing through two polaroids with pass-axes at \( 30^\circ \), if the initial unpo

**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. After the first polaroid, I = (I₀/2) . After the second at 30° , I = (I₀/2) cos² 30° = (I₀/2) × (3/4) = (3I₀/8) . Using Δ = d sinθ, y = n λ D/d, a sinθ = n λ, I = I₀

Ref: NCERT > Physics Book > Wave Optics > Polarization and Malus Law

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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' =

Ref: NCERT > Physics Book > Wave Optics > Polarization and Malus Law

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Polarization and Malus Law

Why does the intensity of light transmitted through three polaroids reach a maximum when the middle one is at 45° to the

**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. At 45°, the middle polaroid allows maximum projection of the polarized light from the first to pass through the third, optimizing the cosine-squared term in Malus’ law. 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 > Polarization and Malus Law

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