How much polarized light passes through a polarizer? Solve I = I₀·cos²θ
three ways — transmitted intensity, incident intensity, or angle — chain up to 6 polarizers
with true relative angles, and watch transmission live on the plot.
I = I₀ · cos² θ
Three-way solver for I = I₀·cos²θ. Pick what to solve for.
Transmitted intensity I
—
Transmission
—
fraction cos²θ as %
Angle
—
degrees = radians
θ=60°, T=25%
Chain up to 6 polarizers. Enter each polarizer’s absolute axis angle;
the tool uses the relative angle between consecutive axes
(light leaves each polarizer polarized along its axis).
Final intensity
—
Overall transmission
—
final I / incident I₀
#
Axis
Relative Δθ
Factor cos²
I after
Quick reference: transmission vs angle
θ
cos²θ
Transmitted (of I₀)
0°
1.0000
100%
15°
0.9330
93.30%
30°
0.7500
75%
45°
0.5000
50%
60°
0.2500
25%
75°
0.0670
6.70%
90°
0.0000
0%
Scope & limits
Ideal polarizers. Real polarizers leak slightly (a typical Polaroid sheet manages about
1000:1 extinction), so crossed real polarizers pass ~0.1% instead of true zero.
Polarized or unpolarized input only. Partially polarized light (degree of polarization
between 0 and 1) is not modelled.
Relative angles in chains. For consecutive polarizers the transmitted factor is
cos² of the relative angle between axes — light leaves each polarizer polarized
along that polarizer’s axis. The tool computes this from your absolute axis inputs.
No absorption or wavelength effects. Real polarizers absorb part of the beam and their
performance varies with wavelength; this calculator models the ideal geometric law only.
Malus's law states that the intensity I of polarized light transmitted through a polarizer is I = I₀ × cos² θ, where I₀ is the incident intensity and θ is the angle between the incoming light's polarization direction and the polarizer's transmission axis. It was discovered by the French physicist Étienne-Louis Malus in 1809.
What is the Malus law formula with an example?
The formula is I = I₀ × cos² θ. For example, with I₀ = 100 W/m² and θ = 60°: cos(60°) = 0.5, so I = 100 × 0.25 = 25 W/m². One quarter of the light gets through.
How much light passes at 0°, 30°, 45°, 60°, and 90°?
With I₀ = 100 W/m²: at 0° the full 100 W/m² passes; at 30°, 75 W/m² (75%); at 45°, 50 W/m² (50%); at 60°, 25 W/m² (25%); at 90°, 0 W/m². Crossed polarizers (90° apart) block everything, for an ideal polarizer.
Why does unpolarized light lose half its intensity at the first polarizer?
Unpolarized light contains all polarization directions equally. Averaging cos² θ over every possible angle gives exactly 1/2, so the first polarizer always transmits I₀/2 — for example 100 W/m² of unpolarized light becomes 50 W/m², now polarized along the first polarizer's axis. Tick the unpolarized box in the multi-polarizer tab and the tool applies this automatically.
How do I calculate intensity through two polarizers?
Use the angle RELATIVE to the previous stage, because light leaving a polarizer is polarized along that polarizer's axis. Example: polarized light (axis 0°) of 100 W/m² passes polarizers at 30° then 45°. Stage 1: relative angle 30°, factor cos² 30° = 0.75, I = 75 W/m². Stage 2: relative angle 15°, factor cos² 15° ≈ 0.9330, I ≈ 69.98 W/m². The multi-polarizer tab above shows every stage's relative angle and factor.
What is the three-polarizer paradox?
Two crossed polarizers (axes 0° and 90°) transmit nothing. Insert a third polarizer at 45° between them and light reappears: 100 W/m² of unpolarized light becomes 50 after the first, 50 × cos² 45° = 25 after the middle one, and 25 × cos² 45° = 12.5 W/m² after the last. The middle polarizer re-orients the polarization so the final stage is only 45° away, not 90°. Try axes 0, 45, 90 in the multi-polarizer tab.
How do I find the angle from two intensity measurements?
Invert the law: θ = arccos(√(I/I₀)). With I = 25 W/m² and I₀ = 100 W/m², I/I₀ = 0.25, √0.25 = 0.5, arccos(0.5) = 60°. Use the “Solve θ” mode above. Note the answer is the principal value between 0° and 90°, because cos² θ is symmetric: θ, −θ, and 180° − θ all give the same intensity.
Which units can I use?
Intensity in W/m² or mW/cm² (1 mW/cm² = 10 W/m²), and angles in degrees or radians (60° = π/3 rad). The tool converts internally and shows the main result in both intensity units.
What are the limitations of this calculator?
It assumes ideal polarizers with perfect extinction. Real polarizers leak a little — a typical Polaroid sheet has an extinction ratio around 1000:1, so crossed real polarizers pass about 0.1% instead of true zero. It also assumes the incident light is fully polarized or fully unpolarized (partial polarization is not modelled), and ignores absorption and wavelength dependence.
How is Malus's law different from Brewster's angle?
Brewster's angle is the angle of incidence at which reflected light becomes perfectly polarized (θ_B = arctan(n₂/n₁); for air to crown glass about 56.65°). Malus's law instead describes how much light passes THROUGH a polarizer once the light is already polarized. See the Brewster angle calculator linked below for the reflection side of polarization.
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