Calculator
Example data table
Example assumes already polarized light with I0 = 100 (arbitrary units).
| Angle θ (deg) | cos²(θ) | Transmitted I |
|---|---|---|
| 0 | 1.000000 | 100.000 |
| 15 | 0.933013 | 93.301 |
| 30 | 0.750000 | 75.000 |
| 45 | 0.500000 | 50.000 |
| 60 | 0.250000 | 25.000 |
| 75 | 0.066987 | 6.699 |
| 90 | 3.749399e-33 | 3.749399e-31 |
Formula used
Malus' law gives transmitted intensity through an ideal analyzer: I = Ip · cos²(θ).
- I is the transmitted intensity after the analyzer.
- Ip is the intensity of polarized light incident on the analyzer.
- θ is the angle between polarization direction and analyzer axis.
If the source is unpolarized and passes one ideal polarizer first, then Ip = I0 / 2.
How to use this calculator
- Select what you want to solve for: I, θ, or I0.
- Choose whether the input is already polarized or unpolarized.
- Enter the required values and select angle units.
- Press Calculate to show results above the form.
- Use the CSV or PDF buttons to export the computed report.
Malus' law in practice
1) What the law predicts
Malus' law relates transmitted intensity after an ideal analyzer to the angle between polarization directions. For a linearly polarized beam, the calculator uses I = Ip·cos²(θ), where Ip is the polarized intensity incident on the analyzer.
2) Key angle benchmarks you can verify
Because the dependence is cos², a few angles become quick checkpoints. At 0°, cos²(0)=1, so I=Ip. At 45°, cos²(45)=0.5, so transmission is 50% of Ip. At 60°, cos²(60)=0.25. At 90°, transmission ideally drops to zero (extinction).
For quick numerical checks, try I0=100 with polarized input. At 15°, cos²(15°)≈0.933 so I≈93.3. At 75°, cos²(75°)≈0.067 so I≈6.7. These match the trend in the example table and help spot entry mistakes.
3) Polarized vs unpolarized input
If the source is already polarized, take Ip = I0 and the analyzer alone sets the output. If the source is unpolarized and passes an ideal polarizer first, the polarized intensity becomes Ip = I0/2. This is why the same analyzer angle can yield half the output when the input is unpolarized.
Example: set unpolarized input with I0=200 and θ=30°. The first polarizer gives Ip=100, then I=100·cos²(30°)=75. Switching to already polarized would double the output to 150 for the same angle.
4) Units and conversion details
The trigonometric part of Malus' law requires radians internally. When you select degrees, the calculator converts using θ(rad) = θ(deg)·π/180. This matters when you copy values from lab notebooks or software that reports angles in radians.
5) Solving for transmitted intensity (I)
In the most common mode, you enter I0 and θ. The report shows cos(θ), cos²(θ), and both transmission ratios: relative to Ip (analyzer only) and relative to I0 (including any unpolarized-to-polarized loss).
6) Solving for angle (θ) from measured intensities
When you provide I0 and measured I, the calculator finds cos²(θ)=I/Ip and reports the principal solution θ=acos(√(I/Ip)). In experiments, remember that θ and 180°−θ give the same cos².
7) Solving for initial intensity (I0) and edge cases
Rearranging gives I0 = I / (pol_factor·cos²(θ)), where pol_factor is 1 (already polarized) or 0.5 (unpolarized with a first polarizer). Near 90°, cos² becomes extremely small, so the computed I0 can blow up and magnify measurement noise.
Use the example table as a sanity check: with I0=100 and polarized input, you should see outputs near 100, 75.0, 25.0, and 0 at 0°, 30°, 60°, and 90° respectively. Small differences come from rounding.
FAQs
1) What is Malus' law used for?
It predicts how much light passes through an analyzer when the light is linearly polarized. It is widely used in optics labs, LCD modeling, polarization microscopy, and simple alignment checks.
2) Why does unpolarized input lose half the intensity?
An ideal polarizer transmits only the field component along its transmission axis. For randomly oriented polarization directions, the average transmitted power is one half of the incident power.
3) Why can two different angles give the same output?
Because cos²(θ) is symmetric: cos²(θ)=cos²(180°−θ). The calculator reports the principal angle, but your physical setup may correspond to the supplementary angle.
4) Does Malus' law work for circular or elliptical polarization?
Not directly in its simplest form. You must first express the light as components along the analyzer axis. Elliptical states require additional parameters, often handled with Stokes vectors and Mueller matrices.
5) What if my polarizers are not ideal?
Real devices have a finite extinction ratio, so the minimum intensity at 90° is not truly zero. Scattering, depolarization, and wavelength effects can also shift results from the ideal curve.
6) Which intensity unit should I pick?
Choose the unit that matches your measurement or simulation output. The math is unit‑agnostic, so W/m², lux, or arbitrary units work the same as long as I0 and I share the unit.
7) Why is solving for I0 unstable near 90°?
At angles near 90°, cos²(θ) approaches zero, so dividing by it amplifies small measurement errors. Use angles away from extinction if you need a reliable back‑calculation of I0.