1. We need the period of the function $f(\theta)=\sin^2\theta.$

2. Use the identity $\sin^2\theta=\frac{1-\cos 2\theta}{2}.$

3. Now analyze the period:

   • The function $\cos x$ has period $2\pi$.
   • Therefore, $\cos 2\theta$ has period $\pi$, because $\cos 2(\theta+\pi)=\cos(2\theta+2\pi)=\cos 2\theta.$

4. Hence, $\sin^2\theta=\frac{1-\cos 2\theta}{2}$ also has period $\pi$.

5. Verify directly: $\sin^2(\theta+\pi)=(-\sin\theta)^2=\sin^2\theta.$ So $\pi$ is a period.

6. Check if any smaller option works: $\sin^2\left(\theta+\frac{\pi}{2}\right)=\cos^2\theta,$ which is not always equal to $\sin^2\theta$. So $\pi/2$ is not the period.

Therefore, the period is $\boxed{\pi}.$ So the correct option is B.