The intensity of transmitted light when a polaroid sheet, placed between two crossed polaroids at 22.5° from the polarization axis of one of the polaroids, is (I₀ is the intensity of polarised light after passing through the first polaroid):

When light passes through a sequence of ideal linear polarizers, the intensity after each polarizer is given by Malus’s law:

$I_1=I_0{cos}^2\left(\theta \right)$

Here I0 is the intensity after the first polarizer, and θ is the angle between the transmission axes of the two successive polarizers.

In the problem we have three polarizers:

• The first polarizer defines the initial polarization.
• The middle polarizer is rotated by 22.5° relative to the first one.
• The third polarizer (the analyzer) is crossed with the first, i.e., its axis is 90° from the first.

Therefore the angles we need are:

$\mathrm{\theta ₁}=\mathrm{22.5°}$

$\mathrm{\theta ₂}=\mathrm{90°}-\mathrm{22.5°}=\mathrm{67.5°}$

Apply Malus’s law twice:

$I_1=I_0{cos}^2\left(\mathrm{22.5°}\right)$

$I_2=I_1{cos}^2\left(\mathrm{67.5°}\right)$

We need the exact values of cos2 at these angles. Using the half‑angle identities:

${cos}^2$

so

${cos}^2$

Similarly,

${cos}^2$

Now multiply the two factors:

$I_2=I_0\times \frac{2+\sqrt{2}}{4}\times \frac{2-\sqrt{2}}{4}$

Notice that the numerators form a difference of squares:

$(2\backslash !+\backslash !\backslash sqrt\{2\})(2\backslash !-\backslash !\backslash sqrt\{2\})\; =\; 4\; -\; 2\; =\; 2$

Hence

$I_2=I_0\backslash times\; \backslash frac\{2\}\{16\}\; =\; \backslash frac\{I\_0\}\{8\}$

Therefore, the intensity of the transmitted light after the three‑polarizer arrangement is

$\frac{I_0}{8}$