In a Young's double-slit experiment, the slit separation is 0.5 mm and the screen is placed 1.0 m away. If the wavelength of light used is 600 nm, what is the fringe width (separation between consecutive bright fringes)?

The fringe width is given by $\beta = \frac{\lambda D}{d}$, where $\lambda = 600 \times 10^{-9}$ m, D = 1.0 m, and d = 0.5 × 10⁻³ m. Thus $\beta = \frac{600 \times 10^{-9} \times 1.0}{0.5 \times 10^{-3}} = \frac{6 \times 10^{-7}}{5 \times 10^{-4}} = 1.2 \times 10^{-3}$ m = 1.20 mm. Distractor A results from omitting the factor of D, while B and D arise from errors in unit conversion.