A diffraction grating has $500$ lines per millimetre and is illuminated by monochromatic light of wavelength $550,\text{nm}$. What is the angle $\theta$ (in degrees) at which the third order ($m=3$) maximum is observed?

Step-by-step solution

1. First, determine the line spacing $d$. Line density $n=500,\text{lines/mm}=5\times 10^5,\text{lines/m}$, so $d=\frac{1}{n}=2.0\times 10^{-6},\text{m}$.
2. Use the grating equation for normal incidence: $sin\theta =mdfrac\lambda d$.
3. Insert $m=3$, $\lambda =550,\text{nm}=5.5\times 10^{-7},\text{m}$, and $d=2.0\times 10^{-6},\text{m}$. Thus $sin\theta =3\times \frac{5.5\times 10^{-7}}{2.0\times 10^{-6}}=0.825$. Take $\theta =\arcsin (0.825)\approx 55.7^{\circ }$. Rounded to nearest degree, $\theta \approx 55^{\circ }$.

Concept tested: Grating diffraction angle

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