In a Young's double slit experiment, the slits are separated by 0.2 mm. If the slits separation is increased to 0.4 mm, the percentage change of the fringe width is:

In Young’s double-slit experiment, the fringe width $$\beta$$ is given by the standard formula

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

where
$$\lambda$$ = wavelength of light,
$$D$$ = distance between the slits and the screen,
$$d$$ = separation between the two slits.

Case 1: Initial slit separation $$d_1 = 0.2\ \text{mm}$$.
Fringe width $$\beta_1 = \frac{\lambda D}{d_1}$$ $$-(1)$$

Case 2: New slit separation $$d_2 = 0.4\ \text{mm} = 2d_1$$.
Fringe width $$\beta_2 = \frac{\lambda D}{d_2} = \frac{\lambda D}{2d_1} = \frac{1}{2}\left(\frac{\lambda D}{d_1}\right)$$

Using $$-(1)$$, $$\beta_2 = \frac{1}{2}\beta_1$$ $$-(2)$$

The percentage change in fringe width is

$$\frac{\beta_2 - \beta_1}{\beta_1}\times 100\% = \frac{\frac{1}{2}\beta_1 - \beta_1}{\beta_1}\times 100\%$$

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

The negative sign shows a decrease. Hence, the fringe width decreases by $$50\%$$.

Therefore, the percentage change (magnitude) is $$50\%$$  ⇒  Option C.