Diffraction Grating Problem Setup
📌 The problem involves two distinct wavelengths of light ($\lambda_1 = 420 \text{ nm}$ for blue and $\lambda_2 = 630 \text{ nm}$ for orange) diffracting from the same grating, overlapping at an angle $\theta = 31^\circ$ from the normal incidence.
📐 The fundamental equation for diffraction maxima is $n\lambda = d \sin\theta$, where $d$ is the line spacing of the grating.
💡 Since the grating ($d$) and the angle ($\theta$) are the same for both wavelengths, the relationship $n_1\lambda_1 = n_2\lambda_2$ must hold true for the overlapping maxima orders ($n_1$ and $n_2$).

Determining the Order Ratio ($n_1/n_2$)
🔗 Rearranging the constant relationship yields the ratio: $\frac{n_1}{n_2} = \frac{\lambda_2}{\lambda_1}$.
🧮 Plugging in the wavelengths gives the ratio $\frac{630 \text{ nm}}{420 \text{ nm}}$, which simplifies exactly to $\frac{3}{2}$.
🔑 Since the order of diffraction ($n$) must be a whole number (e.g., 1, 2, 3...), the smallest acceptable combination that satisfies $n_1/n_2 = 3/2$ is $n_1 = 3$ (for blue light) and $n_2 = 2$ (for orange light).

Calculating the Grating Line Spacing ($d$)
🧪 Using the established orders ($n_2 = 2$) and the orange light wavelength ($\lambda_2 = 630 \times 10^{-9} \text{ m}$), the line spacing $d$ can be calculated using $d = \frac{n_2\lambda_2}{\sin\theta}$.
🧮 Substituting the values: $d = \frac{2 \times (630 \times 10^{-9} \text{ m})}{\sin(31^\circ)}$.
📏 The calculated result is $d \approx 2.446 \times 10^{-6} \text{ m}$, which rounds to $2.4 \times 10^{-6} \text{ m}$ (or $2.4\ \mu\text{m}$), matching the expected multiple-choice answer.

Key Points & Insights
➡️ When facing multiple unknowns in physics problems, identify constants across the different conditions (here, $d$ and $\theta$) to establish an initial solvable relationship ($n_1\lambda_1 = n_2\lambda_2$).
➡️ The constraint that the order of diffraction ($n$) must be a positive integer is crucial for solving ratio problems involving unknown orders.
➡️ In problems where multiple integer solutions exist (e.g., $n_1=6, n_2=4$), always test the smallest valid combination first, as this often corresponds to the solution provided in typical physics problems.
➡️ Double-check all calculations, as a slight error (like using $30^\circ$ instead of $31^\circ$) can lead to a slightly different numerical result, requiring confirmation with the other known variable set (e.g., using the $\lambda_1$ equation).

📸 Video summarized with SummaryTube.com on Jan 17, 2026, 16:19 UTC