A diffraction grating has $N=5000$ lines per centimetre. Monochromatic light of wavelength $\lambda =480\text{nm}$ falls normally on the grating. What is the angle $\theta$ (in degrees) at which the second order ($m=2$) maximum is observed?

Step-by-step solution

1. Find the grating spacing $d$ using $d=\frac{1}{N}$, where $N$ is lines per metre.
2. Convert $N$ to lines per metre: $N=5000{\text{cm}}^{-1}=5.0\times 10^5{\text{m}}^{-1}$.
3. Thus $d=\frac{1}{5.0\times 10^5}=2.0\times 10^{-6}\text{m}$.
4. Use the grating equation $d\sin \theta =m\lambda$ for $m=2$.
5. Substitute: $2.0\times 10^{-6}\sin \theta =2\times 480\times 10^{-9}$.
6. Calculate the right side: $2\times 480\times 10^{-9}=9.6\times 10^{-7}$.
7. Solve for $\sin \theta$: $\sin \theta =\frac{9.6\times 10^{-7}}{2.0\times 10^{-6}}=0.48$.
8. Find $\theta$: $\theta ={\sin }^{-1}(0.48)\approx 28.7^{\circ }$ (rounded to one decimal place).
9. Since the required answer is to one decimal, $\theta \approx 28.7^{\circ }$.

Concept tested: Diffraction grating second‑order angle

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