A single slit of width $a=0.03\text{mm}$ is illuminated by monochromatic light of wavelength $\lambda =550\text{nm}$. Find the smallest angle $\theta$ (in degrees) at which a minimum in the diffraction pattern occurs.

Step-by-step solution

1. For a single‑slit, the first minimum satisfies $a\sin \theta =\lambda$.
2. Convert the width to metres: $a=0.03\text{mm}=3.0\times 10^{-5}\text{m}$.
3. Convert the wavelength: $\lambda =550\text{nm}=5.5\times 10^{-7}\text{m}$.
4. Solve for $\sin \theta$: $\sin \theta =\frac{\lambda }{a}=\frac{5.5\times 10^{-7}}{3.0\times 10^{-5}}=0.01833$.
5. Take the inverse sine: $\theta ={\sin }^{-1}(0.01833)\approx 1.05\text{rad}$.
6. Convert radians to degrees: $1.05\text{rad}\times \frac{180^{\circ }}{\pi }\approx 10.6^{\circ }$.

Concept tested: Single‑slit diffraction minima

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