An unpolarised light beam of intensity I_0 is incident on two...

When unpolarised light of intensity $I_0$ passes through the first polarising sheet, the transmitted light becomes linearly polarised and its intensity is halved:

$I_1 = \dfrac{I_0}{2}$

This linearly polarised light then passes through the second polarising sheet (analyser). According to Malus's law, the final transmitted intensity $I_2$ is:

$I_2 = I_1 \cos^2 \theta$

Given the angle between the transmission axes is $\theta = 45^{\circ}$:

$I_2 = \left(\dfrac{I_0}{2}\right) \cos^2(45^{\circ})$

$I_2 = \left(\dfrac{I_0}{2}\right) \left(\dfrac{1}{\sqrt{2}}\right)^2$

$I_2 = \left(\dfrac{I_0}{2}\right) \left(\dfrac{1}{2}\right) = \dfrac{I_0}{4}$

Forgetting the initial halving of intensity for unpolarised light leads to the incorrect answer $\dfrac{I_0}{2}$.

Answer: $\dfrac{I_0}{4}$