1. Intensity formula in YDSE

For two coherent sources of equal intensity, the intensity at a point with phase difference $\phi$ is

$I = I_{\max} \cos^2\left(\frac{\phi}{2}\right)$

Also, phase difference and path difference $\Delta$ are related by

$\phi = \frac{2\pi}{\lambda}\,\Delta$

So,

$I = I_{\max} \cos^2\left(\frac{\pi \Delta}{\lambda}\right)$

2. Condition for half the peak intensity

We need

$I = \frac{I_{\max}}{2}$

Hence,

$I_{\max} \cos^2\left(\frac{\pi \Delta}{\lambda}\right) = \frac{I_{\max}}{2}$

$\cos^2\left(\frac{\pi \Delta}{\lambda}\right) = \frac{1}{2}$

This gives

$\cos\left(\frac{\pi \Delta}{\lambda}\right) = \pm \frac{1}{\sqrt{2}}$

Therefore,

$\frac{\pi \Delta}{\lambda} = \frac{(2n+1)\pi}{4}, \quad n\in \mathbb{Z}$

So,

$\Delta = \frac{(2n+1)\lambda}{4}$

3. Match with options

The required path difference is

$\boxed{\frac{(2n+1)\lambda}{4}}$

This corresponds to Option B.

4. Comparison with stored answer

Stored correct answer: B

Our derived answer: B

So, they agree.