1. **State the problem:** We have a Young's double-slit experiment with slit separation $d = 0.12$ mm, screen distance $D = 0.80$ m, and the third bright fringe ($m=3$) is displaced $y = 9.0$ mm from the central maximum. We need to find the wavelength $\lambda$ of the light used. 2. **Formula used:** The position of the $m^{th}$ bright fringe in Young's double-slit experiment is given by: $$ y = \frac{m \lambda D}{d} $$ where - $y$ is the fringe displacement from the central maximum, - $m$ is the fringe order (3 in this case), - $\lambda$ is the wavelength, - $D$ is the distance to the screen, - $d$ is the slit separation. 3. **Rearrange the formula to solve for $\lambda$:** $$ \lambda = \frac{y d}{m D} $$ 4. **Convert all units to meters:** - $d = 0.12$ mm = $0.12 \times 10^{-3}$ m = $1.2 \times 10^{-4}$ m - $y = 9.0$ mm = $9.0 \times 10^{-3}$ m - $D = 0.80$ m (already in meters) 5. **Substitute values:** $$ \lambda = \frac{(9.0 \times 10^{-3}) \times (1.2 \times 10^{-4})}{3 \times 0.80} $$ 6. **Calculate numerator and denominator:** $$ \lambda = \frac{1.08 \times 10^{-6}}{2.4} $$ 7. **Simplify the fraction:** $$ \lambda = 4.5 \times 10^{-7} \text{ m} $$ 8. **Final answer:** The wavelength of the light used is $$ \boxed{4.5 \times 10^{-7} \text{ meters} \text{ or } 450 \text{ nm}} $$ This wavelength corresponds to visible light in the blue-green region.