In a Young's double slit experiment two slits are separated by 2 mm and the screen is placed one meter away. When a light of wavelength 500 nm is used, the fringe separation will be:

In Young's double slit experiment, the fringe separation (fringe width) 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, and $$d$$ is the separation between the two slits.

Substituting the given values: $$\lambda = 500 \text{ nm} = 500 \times 10^{-9} \text{ m}$$, $$D = 1 \text{ m}$$, and $$d = 2 \text{ mm} = 2 \times 10^{-3} \text{ m}$$.

$$\beta = \frac{500 \times 10^{-9} \times 1}{2 \times 10^{-3}} = \frac{500 \times 10^{-9}}{2 \times 10^{-3}} = 250 \times 10^{-6} \text{ m} = 0.25 \text{ mm}$$.

Hence the correct answer is Option 4: 0.25 mm.