In a Young’s double‑slit experiment the slit separation is $0.30,\text{mm}$, the wavelength of light used is $550,\text{nm}$, and the screen is placed $2.0,\text{m}$ from the slits. What is the fringe spacing on the screen?

Step-by-step solution

1. "The fringe spacing is given by $\beta =\frac{\lambda L}{d}$ where $L$ is the screen distance and $d$ is the slit separation.
2. "Convert $d=0.30,\text{mm}=3.0\times 10^{-4},\text{m}$, $\lambda =550,\text{nm}=5.5\times 10^{-7},\text{m}$, $L=2.0,\text{m}$.
3. "Compute $\beta =\frac{(5.5\times 10^{-7},\text{m})(2.0,\text{m})}{3.0\times 10^{-4},\text{m}}=3.6667\times 10^{-3},\text{m}$.
4. "Convert to millimetres: $3.6667\times 10^{-3},\text{m}=3.67,\text{mm}$.

Concept tested: Young’s double‑slit interference

Published 16 September 2026 · How this solution was prepared and checked