In a Young's double slit experiment, the line-width of a bright...

The intensity of light in Young's double slit experiment is given by I = I_\max \cos^2 \left(\dfrac{\phi}{2} \right), where $\phi$ is the phase difference.

For the intensity to drop to half of its maximum value, $I =$\dfrac{I_\max}{2}.

\cos^2 \left(\dfrac{\phi}{2} \right) = \dfrac{1}{2} \Rightarrow \cos \left( \dfrac{\phi}{2} \right) = \pm \dfrac{1}{\sqrt{2}}

\dfrac{\phi}{2} = \pm \dfrac{\pi}{4} \Rightarrow \phi = \pm \dfrac{\pi}{2}

The phase difference $\phi$ is related to the path difference $\Delta x$ by $\phi =$\dfrac{2\pi}{\lambda} \Delta$x$.

\dfrac{2\pi}{\lambda} \Delta$x = \pm$\dfrac{\pi}{2} \Rightarrow \Delta$x = \pm$\dfrac{\lambda}{4}

In YDSE, the path difference is $\Delta x =$\dfrac{y d}{D}, where $y$ is the distance from the central maximum.

\dfrac{y d}{D} = \pm \dfrac{\lambda}{4} \Rightarrow$y = \pm$\dfrac{D\lambda}{4d}

The line-width is the distance between these two points on either side of the central maximum:

$\Delta y =$\dfrac{D\lambda}{4d} - \left( - \dfrac{D\lambda}{4d} \right) = \dfrac{D\lambda}{2d}