In Young's double slit experiment, the fringe width is $$12 \text{ mm}$$. If the entire arrangement is placed in water of refractive index $$\dfrac{4}{3}$$, then the fringe width becomes (in mm)

In Young's double slit experiment, the fringe width in air is $$\beta = 12 \text{ mm}$$. The entire arrangement is placed in water of refractive index $$\mu = \dfrac{4}{3}$$. We need to find the new fringe width.

The fringe width in Young's double slit experiment is given by $$\beta = \frac{\lambda D}{d}$$, where $$\lambda$$ is the wavelength, $$D$$ is the distance to the screen, and $$d$$ is the slit separation.

Since the setup is placed in a medium of refractive index $$\mu$$, the wavelength changes to $$\lambda' = \frac{\lambda}{\mu}$$ while the distances $$D$$ and $$d$$ remain unchanged.

Next, the new fringe width is $$\beta' = \frac{\lambda' D}{d} = \frac{\lambda D}{\mu d} = \frac{\beta}{\mu}$$. Substituting gives $$\beta' = \frac{12}{4/3} = 12 \times \frac{3}{4} = 9 \text{ mm}$$.

The correct answer is Option B: $$9$$ mm.