Problem 5.64 — Diffraction Grating: Principal Maxima

Problem Statement

A diffraction grating has $N = 500$ lines/mm. Find the angles of the first three principal maxima for $\lambda = 600$ nm.

Given Information

  • See problem statement for all given quantities.

Physical Concepts & Formulas

This problem applies fundamental physics principles to the scenario described. The solution requires identifying the relevant conservation laws and equations of motion, then solving systematically with careful attention to units and sign conventions.

  • See the step-by-step solution for the specific equations applied.
  • All quantities are in SI units unless otherwise stated.

Step-by-Step Solution

Step 1 — Identify given quantities and set up the problem: Grating equation: $d\sin\theta = m\lambda$, where $d = 1/500$ mm $= 2.0\times10^{-6}$ m.

Step 2 — Apply the relevant physical law or equation: $m = 1$: $\sin\theta_1 = 600\times10^{-9}/2\times10^{-6} = 0.300 \implies \theta_1 = 17.5°$

Step 3 — Solve algebraically for the unknown: $m = 2$: $\sin\theta_2 = 0.600 \implies \theta_2 = 36.9°$

Step 4 — Substitute numerical values with units: $m = 3$: $\sin\theta_3 = 0.900 \implies \theta_3 = 64.2°$

Worked Calculation

$$\text{Numerical result} = \text{given expression substituted with values}$$

$$\boxed{\sin\theta_3 = 0.900 \implies \theta_3 = 64.2°}$$

Grating equation: $d\sin\theta = m\lambda$, where $d = 1/500$ mm $= 2.0\times10^{-6}$ m.

$m = 1$: $\sin\theta_1 = 600\times10^{-9}/2\times10^{-6} = 0.300 \implies \theta_1 = 17.5°$

$m = 2$: $\sin\theta_2 = 0.600 \implies \theta_2 = 36.9°$

$m = 3$: $\sin\theta_3 = 0.900 \implies \theta_3 = 64.2°$

Answer

$$\boxed{\sin\theta_3 = 0.900 \implies \theta_3 = 64.2°}$$

Physical Interpretation

The numerical answer is physically reasonable — matching expected orders of magnitude and dimensional analysis. The result confirms the theoretical prediction and provides quantitative insight into the system’s behaviour.


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