1. Understand the setup

   • Two polaroids are crossed, so their transmission axes are at $90^\circ$ to each other.
   • A third polaroid sheet is inserted between them and rotated by some angle $\theta$ with respect to the first polaroid.

2. Apply Malus' law step-by-step

   Let the intensity of unpolarized incident light be $I_0$.

   • After passing through the first polaroid: $I_1 = \frac{I_0}{2}$

   • The middle polaroid is at angle $\theta$ relative to the first, so after passing through it: $I_2 = I_1 \cos^2\theta = \frac{I_0}{2}\cos^2\theta$

   • The third polaroid is crossed with the first, so it is at angle $90^\circ$ to the first. Hence the angle between the middle and the third polaroid is: $90^\circ - \theta$

   Therefore, final transmitted intensity is: $I = I_2 \cos^2(90^\circ - \theta)$

   Using $\cos(90^\circ-\theta)=\sin\theta$, $I = \frac{I_0}{2}\cos^2\theta\sin^2\theta$

3. Simplify the expression

   $I = \frac{I_0}{2}\cos^2\theta\sin^2\theta$

   Using $\sin^2 2\theta = 4\sin^2\theta\cos^2\theta$

   we get: $I = \frac{I_0}{8}\sin^2 2\theta$

4. Find maximum intensity

   The intensity is maximum when $\sin^2 2\theta = 1$ which happens when $2\theta = 90^\circ$ so, $\theta = 45^\circ$

5. Check options

   • A: $90^\circ$ → not maximum
   • B: $30^\circ$ → not maximum
   • C: $45^\circ$ → correct
   • D: $60^\circ$ → not maximum

Thus, the transmitted light intensity is maximum when the middle polaroid is rotated by $45^\circ$.