A single slit of width $a=0.04\text{mm}$ is illuminated by monochromatic light of wavelength $\lambda =600\text{nm}$. The screen is placed $D=2.0\text{m}$ behind the slit. Determine the linear width $w$ of the central bright fringe on the screen.

Step-by-step solution

1. The first minima occur at angles satisfying $a\sin \theta =\pm \lambda$.
2. For small angles, $\sin \theta \approx \tan \theta =\frac{y}{D}$, where $y$ is the distance from the central maximum to the first minimum on the screen.
3. Thus $a\frac{y}{D}=\lambda \Rightarrow y=\frac{\lambda D}{a}$.
4. Convert units: $a=0.04\text{mm}=4.0\times 10^{-5}\text{m}$, $\lambda =600\text{nm}=6.0\times 10^{-7}\text{m}$, $D=2.0\text{m}$.
5. Compute $y$: $y=\frac{6.0\times 10^{-7}\times 2.0}{4.0\times 10^{-5}}=\frac{1.2\times 10^{-6}}{4.0\times 10^{-5}}=0.03\text{m}=30\text{mm}$.
6. The central bright fringe extends from $-y$ to $+y$, so its total width $w=2y=2\times 30\text{mm}=60\text{mm}$.
7. Check the calculation: using the correct small‑angle approximation gives $y=\frac{\lambda D}{a}=\frac{6.0\times 10^{-7}\times 2}{4.0\times 10^{-5}}=0.03\text{m}$, thus $w=0.06\text{m}=60\text{mm}$.
8. However, the problem asks for the width of the central bright *fringe*, which is the distance between the first minima on either side, i.e., $w=2y=0.06\text{m}=60\text{mm}=6.0\text{cm}$.

Concept tested: Single‑slit central maximum width

Published 7 September 2026 · How this solution was prepared and checked