Two polaroids are placed in the path of unpolarised beam of intensity ' $\mathrm{I}_0$ ' such that no light is emitted from the second polaroid. If a third polaroid whose polarisation axis makes an angle ' $\theta$ ' with the polarisation axis of first polaroid is placed between these polaroids then the intensity of light emerging from the last polaroid will be

The given solution describes the intensity of light transmitted through a system of Polaroids. The first Polaroid (P1) is placed with its transmission axis vertical. The second Polaroid (P2) is placed such that its transmission axis is horizontal, making P1 and P2 crossed Polaroids. A third Polaroid (P3) is inserted between them at an angle $\theta$ to the vertical.

When unpolarized light of intensity $I_0$ is incident on P1, the transmitted intensity is $I_1 = \frac{I_0}{2}$.

The light then reaches P3, which is oriented at an angle $\theta$ to P1. By Malus's law, the intensity after P3 is $I_3 = I_1 \cos^2\theta = \frac{I_0}{2} \cos^2\theta$.

Finally, the light passes through P2. The angle between the transmission axis of P3 and P2 is $90^\circ - \theta$. Applying Malus's law again gives $I_2 = I_3 \cos^2(90^\circ - \theta) = I_3 \sin^2\theta$.

Substituting the expression for $I_3$ yields $I_2 = \frac{I_0}{2} \cos^2\theta \sin^2\theta$.

Using the identity $\sin^2\theta \cos^2\theta = \frac{1}{4} \sin^2 2\theta$, the intensity simplifies to $I_2 = \frac{I_0}{8} \sin^2 2\theta$.

Thus, the correct expression for the intensity of light transmitted through the system is $\boxed{\text{B}}$.