Problem Statement
A blazed reflection grating has blaze angle $\gamma = 15°$ and is used in Littrow configuration (reflected light goes back along the incident path). Find the wavelength that is blazed most efficiently for 1st order.
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: In Littrow configuration, $\theta_i = \theta_r = \gamma$. Grating equation: $2d\sin\gamma = m\lambda_{blaze}$.
Step 2 — Apply the relevant physical law or equation: With $d = 1.67\;\mu$m (600 gr/mm grating assumed) and $m = 1$:
Step 3 — Solve algebraically for the unknown: $$\lambda_{blaze} = 2d\sin\gamma = 2\times1.67\times10^{-6}\times\sin15° = 2\times1.67\times10^{-6}\times0.259 = 865\text{ nm}$$
Step 4 — Substitute numerical values with units: (Result depends on grating period; general formula: $\lambda_{blaze} = 2d\sin\gamma/m$.)
Worked Calculation
$$\lambda_{blaze} = 2d\sin\gamma = 2\times1.67\times10^{-6}\times\sin15° = 2\times1.67\times10^{-6}\times0.259 = 865\text{ nm}$$
$$\text{Numerical result} = \text{given expression substituted with values}$$
$$\boxed{\lambda_{blaze} = 2d\sin\gamma = 2\times1.67\times10^{-6}\times\sin15° = 2\times1.67\times10^{-6}\times0.259 = 865\text{ nm}}$$
In Littrow configuration, $\theta_i = \theta_r = \gamma$. Grating equation: $2d\sin\gamma = m\lambda_{blaze}$.
With $d = 1.67\;\mu$m (600 gr/mm grating assumed) and $m = 1$:
$$\lambda_{blaze} = 2d\sin\gamma = 2\times1.67\times10^{-6}\times\sin15° = 2\times1.67\times10^{-6}\times0.259 = 865\text{ nm}$$
(Result depends on grating period; general formula: $\lambda_{blaze} = 2d\sin\gamma/m$.)
Answer
$$\boxed{\lambda_{blaze} = 2d\sin\gamma = 2\times1.67\times10^{-6}\times\sin15° = 2\times1.67\times10^{-6}\times0.259 = 865\text{ nm}}$$
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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