The intensity of transmitted light when a polaroid sheet, placed between two crossed polarization at 22.5° from the polarization axis of one of the polaroid, is (I₀ is the intensity or 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.

Step 1: Understand the setup of the polaroids
We are given two crossed polaroids. Let the first polaroid be $P_1$ and the third polaroid be $P_3$. Since they are crossed, the angle between their transmission axes is $90°$.
The intensity of the polarized light after passing through the first polaroid $P_1$ is given as $I_0$.

Step 2: Introduce the middle polaroid
A third polaroid sheet, $P_2$, is placed between $P_1$ and $P_3$ at an angle of $22.5°$ relative to the transmission axis of $P_1$.
Therefore, the angle between the axes of $P_1$ and $P_2$ is:
${\theta }_1=22.5°$
Since $P_1$ and $P_3$ are crossed ($90°$ relative to each other), the angle between the axes of $P_2$ and $P_3$ must be:
${\theta }_2=90°-22.5°=67.5°$

Step 3: Calculate the intensity after the second polaroid ($P_2$)
According to Malus's Law, the intensity of light $I_1$ transmitted through $P_2$ is:

$I_1=I_0{\cos }^2(22.5°)$

Step 4: Calculate the intensity after the third polaroid ($P_3$)
The light of intensity $I_1$ now passes through $P_3$. Using Malus's Law again, the final transmitted intensity $I$ is:

$I=I_1{\cos }^2(67.5°)$

Substituting the expression for $I_1$:

$I=I_0{\cos }^2(22.5°){\cos }^2(67.5°)$

Step 5: Simplify using trigonometric identities
We use the identity $\cos (90°-x)=\sin \left(x\right)$. Thus:
$\cos (67.5°)=\sin (22.5°)$
Substituting this back into the intensity formula:

$I=I_0{\cos }^2(22.5°){\sin }^2(22.5°)$

We can rewrite this expression as:

$I=I_0{\left[\sin \right(22.5°\left)\cos \right(22.5°\left)\right]}^2$

Using the double-angle identity for sine, $\sin \left(2\theta \right)=2\sin \left(\theta \right)\cos \left(\theta \right)$, we have:

$\sin (22.5°)\cos (22.5°)=\frac{1}{2}\sin (45°)$

Since $\sin (45°)=\frac{1}{\sqrt{2}}$:

$\sin (22.5°)\cos (22.5°)=\frac{1}{2\sqrt{2}}$

Squaring this term gives:

${\left[\sin \right(22.5°\left)\cos \right(22.5°\left)\right]}^2={\left(\frac{1}{2\sqrt{2}}\right)}^2=\frac{1}{8}$

Therefore, the final transmitted intensity is:

$I=\frac{I_0}{8}$