A polaroid sheet is rotated between two crossed polarizers. The intensity of transmitted light would be maximum, when the angle between the axes of the first polarizer and the polaroid sheet is

The correct option is π/4.

Let us understand the setup and derive the transmitted intensity step-by-step using Malus's Law.

We start with two crossed polarizers. This means the transmission axis of the first polarizer ($P_1$) is perpendicular to the transmission axis of the second polarizer ($P_2$). The angle between their axes is $\frac{\pi }{2}$.

Now, a third polaroid sheet ($P_3$) is inserted and rotated between them. Let the angle between the transmission axis of the first polarizer ($P_1$) and the middle sheet ($P_3$) be $\theta$.

Since the first and second polarizers are crossed, the angle between the middle sheet ($P_3$) and the second polarizer ($P_2$) will be:
$\frac{\pi }{2}-\theta$

Let $I_0$ be the intensity of light transmitted by the first polarizer $P_1$. When this light passes through the middle sheet $P_3$, the transmitted intensity $I_1$ is given by Malus's Law:
$I_1=I_0{\cos }^2\theta$

This light of intensity $I_1$ then incident on the second polarizer $P_2$. The final transmitted intensity $I$ is:
$I=I_1{\cos }^2(\frac{\pi }{2}-\theta )$

Using the trigonometric identity $\cos (\frac{\pi }{2}-\theta )=\sin \theta$, we get:
$I=I_1{\sin }^2\theta$

Substituting the value of $I_1$ into the equation:
$I=I_0{\cos }^2\theta {\sin }^2\theta$

We can rewrite this expression by grouping the sine and cosine terms:
$I=I_0{\left(\sin \theta \cos \theta \right)}^2$

Using the double-angle identity $\sin \left(2\theta \right)=2\sin \theta \cos \theta$, we have:
$I=I_0{\left(\frac{\sin \left(2\theta \right)}{2}\right)}^2=\frac{I_0}{4}{\sin }^2\left(2\theta \right)$

To make the transmitted intensity $I$ maximum, the term ${\sin }^2\left(2\theta \right)$ must be at its maximum value, which is 1:
${\sin }^2\left(2\theta \right)=1$
$\sin \left(2\theta \right)=1$

This occurs when:
$2\theta =\frac{\pi }{2}$
$\theta =\frac{\pi }{4}$

Therefore, the transmitted light intensity is maximum when the angle between the axes of the first polarizer and the polaroid sheet is $\frac{\pi }{4}$.