A diffraction grating has $1200$ lines per centimetre and is illuminated by monochromatic light of wavelength $\lambda =500$ nm. (a) Find the angle $\theta$ (in degrees) at which the third order ($m=3$) maximum is observed. (b) A screen is placed $D=4$ m behind the grating. What is the linear separation $s$ between the two third‑order maxima on the screen? (c) Which of the following choices gives the correct value of $s$?

Step-by-step solution

1. (a) Convert the line density to line spacing: $d=\frac{1,\text{cm}}{1200}=8.333\times 10^{-6}\text{ m}$. Apply the grating equation $mlambda=dsin\theta$ for $m=3$: $sin\theta =\frac{3\times 500\times 10^{-9}}{8.333\times 10^{-6}}\approx 0.18$. Hence $\theta =\arcsin (0.18)\approx 10.3^{\circ }$. (b) For small angles, $\tan \theta \approx \sin \theta$. The linear separation on a screen at distance $D$ is $s=D\tan \theta \approx D,sin\theta$. Insert $D=4\text{ m}$ and $sin\theta =0.18$: $sapprox4\times 0.18=0.72\text{ m}$.
2. (c) Options: - $0.55\text{ m}$ - $0.72\text{ m}$ - $0.90\text{ m}$ - $1.10\text{ m}$ Correct choice: $0.72\text{ m}$.

Concept tested: Grating diffraction angle and image separation

Published 4 October 2026 · How this solution was prepared and checked