In Young's double slits experiment, the position of 5$$^{th}$$ bright fringe from the central maximum is 5 cm. The distance between slits and screen is 1 m and wavelength of used monochromatic light is 600 nm. The separation between the slits is:

In Young's double slit experiment the position of the 5th bright fringe from the central maximum is $$y_5 = 5 \text{ cm} = 5 \times 10^{-2} \text{ m}$$, the distance between the slits and the screen is $$D = 1 \text{ m}$$, and the wavelength is $$\lambda = 600 \text{ nm} = 600 \times 10^{-9} \text{ m}$$.

The position of the $$n$$th bright fringe is given by the relation $$y_n = \frac{n \lambda D}{d}$$, where $$d$$ is the separation between the slits.

Solving this relation for $$d$$ gives $$d = \frac{n \lambda D}{y_n} = \frac{5 \times 600 \times 10^{-9} \times 1}{5 \times 10^{-2}}$$. Further simplification leads to $$d = \frac{3000 \times 10^{-9}}{5 \times 10^{-2}} = \frac{3 \times 10^{-6}}{5 \times 10^{-2}} = 6 \times 10^{-5} \text{ m}$$ and finally $$d = 60 \times 10^{-6} \text{ m} = 60 \; \mu\text{m}$$.

The correct answer is Option A: 60 $$\mu$$m.