A single slit of width $0.05,\text{mm}$ is illuminated by monochromatic light of wavelength $650,\text{nm}$ and a screen is placed $2.50,\text{m}$ behind the slit. What is the linear width of the central bright fringe on the screen?

Step-by-step solution

1. The first minima occur when $sin\theta =pmdfrac\lambda a$; for small angles $sin\theta \simeq \theta$ in radians.
2. The distance from the centre to a first minimum on the screen is $y=a,\frac{D}{\lambda }$? No, correct formula: $y=D\tan \theta \simeq D\theta$ with $\theta \simeq \frac{\lambda }{a}$.
3. Thus $y_{\text{min}}\simeq D\frac{\lambda }{a}$. The central maximum extends from $-y_{\text{min}}$ to $+y_{\text{min}}$, so its width is $w=2y_{\text{min}}=2D\frac{\lambda }{a}$.
4. Convert quantities: $a=0.05,\text{mm}=5.0\times 10^{-5},\text{m}$, $\lambda =650,\text{nm}=6.5\times 10^{-7},\text{m}$, $D=2.50,\text{m}$.
5. Compute $w=2\times 2.50\times \frac{6.5\times 10^{-7}}{5.0\times 10^{-5}};\text{m}$.
6. Evaluate numerator: $2\times 2.50\times 6.5\times 10^{-7}=3.25\times 10^{-6};\text{m}$. Divide by $5.0\times 10^{-5}$ to obtain $w=6.50\times 10^{-2},\text{m}$.
7. Convert to millimetres: $6.50\times 10^{-2},\text{m}=65,\text{mm}$.

Concept tested: Single‑slit diffraction central width

Published 21 September 2026 · How this solution was prepared and checked