A student sets up a double-slit experiment…

Step 1. The fringe widths. $\beta_A = \dfrac{600 \times 10^{-9}\ \text{m} \times 1.50\ \text{m}}{0.30 \times 10^{-3}\ \text{m}} = 3.00\ \text{mm}$ and $\beta_B = \dfrac{500 \times 10^{-9}\ \text{m} \times 1.50\ \text{m}}{0.30 \times 10^{-3}\ \text{m}} = 2.50\ \text{mm}$.

Step 2. Two bright fringes meet. We need $n_1\beta_A = n_2\beta_B$. The ratio $\beta_A/\beta_B = 3.00/2.50 = \tfrac{6}{5}$, so $n_1 \times 6 = n_2 \times 5$. The smallest whole numbers are $n_1 = 5$ and $n_2 = 6$. The place is $5 \times 3.00\ \text{mm} = 15.0\ \text{mm}$. They meet again at $30.0$ mm.

Step 3. Bright on dark. A dark fringe of the longer wavelength is at $\left(n_1 + \tfrac12\right) \times 3.00\ \text{mm}$. The first one that meets a bright fringe $n_2 \times 2.50\ \text{mm}$ is at $n_1 = 2$ and $n_2 = 3$: $2.5 \times 3.00\ \text{mm} = 3 \times 2.50\ \text{mm} = 7.5\ \text{mm}$.