In Young's double slit experiment, the fringe width $$\beta$$ is given by:

$$\beta = \frac{\lambda D}{d}$$

where:

$$\lambda$$ is the wavelength of light,

$$D$$ is the distance from the slits to the screen,

$$d$$ is the separation between the slits.

Here's how the change affects the fringe width:

Initial Situation:

When $$d = 0.2 \text{ mm}$$, the fringe width is:

$$\beta_{\text{initial}} = \frac{\lambda D}{0.2 \text{ mm}}$$

After Increasing Slit Separation:

When $$d$$ is increased to $$0.4 \text{ mm}$$:

$$\beta_{\text{new}} = \frac{\lambda D}{0.4 \text{ mm}}$$

Comparing the Two Fringe Widths:

Notice that:

$$\beta_{\text{new}} = \frac{1}{2} \times \frac{\lambda D}{0.2 \text{ mm}} = \frac{1}{2} \beta_{\text{initial}}$$

This means the fringe width is halved, which is a reduction of $$50\%$$.

Thus, the fringe width decreases by $$50\%$$ when the slit separation is increased from $$0.2 \text{ mm}$$ to $$0.4 \text{ mm}$$.

The correct answer is Option B: $$50\%$$.