In a Young’s double‑slit experiment the slit separation is $d=0.25\text{mm}$ and monochromatic light of wavelength $\lambda =480\text{nm}$ is used. The screen is placed $L=1.5\text{m}$ from the slits. Calculate the fringe spacing $\beta$ (distance between two consecutive bright fringes) on the screen.

Step-by-step solution

1. Write the formula for fringe spacing in Young’s experiment: $\beta =\frac{\lambda L}{d}$.
2. Convert all quantities to metres: $d=0.25\text{mm}=2.5\times 10^{-4}\text{m}$, $\lambda =480\text{nm}=4.8\times 10^{-7}\text{m}$, $L=1.5\text{m}$.
3. Substitute the values: $\beta =\frac{4.8\times 10^{-7}\times 1.5}{2.5\times 10^{-4}}$.
4. Perform the multiplication in the numerator: $4.8\times 10^{-7}\times 1.5=7.2\times 10^{-7}$.
5. Divide: $\frac{7.2\times 10^{-7}}{2.5\times 10^{-4}}=2.88\times 10^{-3}\text{m}$.
6. Convert metres to millimetres: $2.88\times 10^{-3}\text{m}=2.88\text{mm}$.

Concept tested: Young's double‑slit interference

Published 10 September 2026 · How this solution was prepared and checked