To find the fringe separation (also known as fringe width) in Young's double slit experiment, we use the formula:

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

where:

$\beta$ is the fringe width (fringe separation).

$\lambda$ is the wavelength of the light used.

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

$d$ is the distance between the two slits.

Given values are:

$$d = 1.5 \, \mathrm{mm} = 1.5 \times 10^{-3} \, \mathrm{m}$$

$D = 1 \, \mathrm{m}$

$$\lambda = 600 \times 10^{-9} \, \mathrm{m}$$

Substituting these values into the formula:

$$ \beta = \dfrac{600 \times 10^{-9} \, \mathrm{m} \times 1 \, \mathrm{m}}{1.5 \times 10^{-3} \, \mathrm{m}} $$

Calculating the value:

$$ \beta = \dfrac{600 \times 10^{-9}}{1.5 \times 10^{-3}} $$

$$ \beta = 4 \times 10^{-4} \, \mathrm{m} $$

Thus, the fringe separation is $$4 \times 10^{-4} \, \mathrm{m}$$. The correct answer is:

Option D $$4 \times 10^{-4} \mathrm{~m}$$