In a Young's double slit experiment with light of wavelength $$\lambda$$, the separation of slits is $$d$$ and distance of screen is $$D$$ such that $$D \gg d \gg \lambda$$. If the Fringe width is $$\beta$$, the distance from point of maximum intensity to the point where intensity falls to half of the maximum intensity on either side is:

We begin with the standard expression for the intensity distribution on the screen in a Young’s double slit experiment. If the two individual slits, kept in phase, each produce an intensity $$I_0,$$ then at a point situated a distance $$y$$ from the central line on the screen the resultant intensity is given by the interference formula

$$I \;=\; 4I_0\cos^2\!\theta,$$

where the phase angle $$\theta$$ is related to the geometry by

$$\theta \;=\;\dfrac{\pi\,d\,y}{\lambda D}.$$

Here $$d$$ is the slit separation, $$D$$ the distance of the screen from the slits, and $$\lambda$$ the wavelength of light. The factor $$4I_0$$ represents the maximum possible intensity when the two waves interfere constructively in perfect phase.

For the central bright fringe, the path difference is zero, so $$\theta = 0$$ and we have

$$I_{\text{max}} \;=\; 4I_0\cos^2 0 \;=\; 4I_0.$$

We are asked to locate the point (on either side of the central maximum) where the intensity falls to half of this maximum value, that is

$$I \;=\;\dfrac{I_{\text{max}}}{2} \;=\;\dfrac{4I_0}{2} \;=\; 2I_0.$$

Substituting $$I = 4I_0\cos^2\!\theta$$ into the above condition gives

$$4I_0\cos^2\!\theta \;=\; 2I_0.$$

Dividing both sides by $$2I_0$$ we get

$$2\cos^2\!\theta \;=\; 1 \;\Longrightarrow\; \cos^2\!\theta \;=\;\dfrac{1}{2}.$$

Taking the square root (and remembering that the cosine function is even) yields

$$\cos\!\theta \;=\;\dfrac{1}{\sqrt{2}}.$$

Since $$\cos 45^{\circ} = 1/\sqrt{2},$$ the smallest positive value of $$\theta$$ satisfying the above is

$$\theta \;=\;\dfrac{\pi}{4}.$$

Now we insert this value back into the geometrical relation for $$\theta$$:

$$\dfrac{\pi\,d\,y}{\lambda D} \;=\;\dfrac{\pi}{4}.$$

We cancel $$\pi$$ from both sides to obtain

$$\dfrac{d\,y}{\lambda D} \;=\;\dfrac{1}{4}.$$

Multiplying through by $$\dfrac{\lambda D}{d}$$ gives the required distance $$y$$ of the half-intensity point from the central maximum:

$$y \;=\;\dfrac{\lambda D}{4d}.$$

Next we recall the definition of the fringe width $$\beta,$$ which is the separation between two successive bright (or dark) fringes. For Young’s experiment it is

$$\beta \;=\;\dfrac{\lambda D}{d}.$$

Substituting this value for $$\dfrac{\lambda D}{d}$$ into the expression for $$y$$ we get

$$y \;=\;\dfrac{\beta}{4}.$$

This distance $$y = \beta/4$$ is measured from the central maximum to the position where the intensity has dropped to one-half of the maximum on either side. Therefore the required distance is $$\dfrac{\beta}{4}.$$

Hence, the correct answer is Option A.