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):

To find the intensity of the transmitted light, let us analyze the setup step-by-step using Malus's Law.

Malus's Law states that when completely plane-polarized light of intensity $I_i$ is incident on an analyzer, the intensity of the light transmitted through the analyzer, $I_t$, is given by:
$I_t=I_i{\cos }^2\left(\theta \right)$
where $\theta$ is the angle between the transmission axis of the polarizer and the analyzer.

Let the three polaroids be P₁, P₂, and P₃:
1. P₁ is the first polaroid. The light passing through P₁ becomes plane-polarized with an intensity of $I_0$.
2. P₃ is the third polaroid. It is crossed with P₁, which means the angle between the transmission axis of P₁ and P₃ is 90°.
3. P₂ is the second polaroid sheet inserted between P₁ and P₃. Its transmission axis is at an angle of ${\theta }_1=\; 22.5°$ with respect to the axis of P₁.

First, we calculate the intensity of light transmitted through P₂ (let this be $I_1$):
$I_1=I_0{\cos }^2\left(\mathrm{22.5°}\right)$

Next, the light transmitted through P₂ is incident on P₃. The angle ${\theta }_2$ between the axis of P₂ and P₃ is:
${\theta }_2=\; 90°\; -\; 22.5°\; =\; 67.5°$

The final transmitted intensity $I$ through P₃ is:
$I=I_1{\cos }^2\left(\mathrm{67.5°}\right)$
Since $\cos \left(\mathrm{67.5°}\right)=\sin \left(\mathrm{22.5°}\right)$, we can write:
$I=I_0{\cos }^2\left(\mathrm{22.5°}\right){\sin }^2\left(\mathrm{22.5°}\right)$

We can rewrite this expression using the trigonometric identity $\sin \left(2\theta \right)=\; 2$ $\sin \left(\theta \right)$ $\cos \left(\theta \right)$:
$I=I_0{\left[\sin \right(\mathrm{22.5°}\left)\cos \right(\mathrm{22.5°}\left)\right]}^2$
$I=I_0{\left[\frac{\sin (2\times \mathrm{22.5°})}{2}\right]}^2$
$I=I_0{\left[\frac{\sin \left(\mathrm{45°}\right)}{2}\right]}^2$

Since $\sin \left(\mathrm{45°}\right)=\frac{1}{\sqrt{2}}$, substituting this value gives:
$I=I_0{\left[\frac{1}{2\sqrt{2}}\right]}^2$
$I=I_0\left(\frac{1}{8}\right)=\frac{I_0}{8}$

Therefore, the intensity of the transmitted light is $I_0/8$.