1. In Young’s double slit experiment, the fringe width is given by

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

where:

• $\lambda$ = wavelength of light
• $D$ = distance of screen from slits
• $d$ = separation between slits

2. Here, initially

$d_1 = 0.2\text{ mm}$

and finally

$d_2 = 0.4\text{ mm}$

3. Since fringe width is inversely proportional to slit separation,

$\beta \propto \frac{1}{d}$

So,

$\frac{\beta_2}{\beta_1} = \frac{d_1}{d_2} = \frac{0.2}{0.4} = \frac{1}{2}$

Thus,

$\beta_2 = \frac{\beta_1}{2}$

So the fringe width becomes half of its initial value.

4. Percentage change in fringe width:

$\text{Percentage change} = \frac{\beta_2 - \beta_1}{\beta_1} \times 100$

$= \frac{\frac{\beta_1}{2} - \beta_1}{\beta_1} \times 100$

$= \left(\frac{-1}{2}\right) \times 100 = -50\%$

The negative sign means a decrease.

So, the fringe width decreases by $50\%$.

5. Therefore, the correct option is:

$\boxed{\text{B: }50\%}$