In Young's double slits experiment, the light has a frequency 12 × 10¹⁴ Hz and the distance between the centers of adjacent fringes is 0.8 mm. If the screen is 1.6 m away, what is the distance between the slits? (Take speed of light 3 × 10⁸ m s^-1)

5.00 × 10^-4 m

$ \lambda = \frac{c}{\nu} = \frac{3\times 10^8}{12\times 10^14} = 250nm$

$ \beta = \frac{\lambda D}{d} \Rightarrow d = \frac{\lambda D}{\beta} = 5\times 10^{-4}$