Problem Statement
A hologram is recorded with object and reference beams from the same laser ($\lambda = 633$ nm) making an angle of $30°$ between them. Find the spatial frequency of the interference fringes recorded in the hologram.
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: The two beams interfere with a fringe spacing:
Step 2 — Apply the relevant physical law or equation: $$\Lambda = \frac{\lambda}{2\sin(\theta/2)} = \frac{633\times10^{-9}}{2\sin15°} = \frac{633\times10^{-9}}{0.518} \approx 1.22\;\mu\text{m}$$
$$f = 1/\Lambda \approx \boxed{820\text{ lines/mm}}$$
Step 3 — Verify the result: Check units, limiting cases, and order of magnitude to confirm the answer is physically reasonable.
Worked Calculation
$$\Lambda = \frac{\lambda}{2\sin(\theta/2)} = \frac{633\times10^{-9}}{2\sin15°} = \frac{633\times10^{-9}}{0.518} \approx 1.22\;\mu\text{m}$$
$$f = 1/\Lambda \approx \boxed{820\text{ lines/mm}}$$
$$\text{Numerical result} = \text{given expression substituted with values}$$
The two beams interfere with a fringe spacing:
$$\Lambda = \frac{\lambda}{2\sin(\theta/2)} = \frac{633\times10^{-9}}{2\sin15°} = \frac{633\times10^{-9}}{0.518} \approx 1.22\;\mu\text{m}$$
$$f = 1/\Lambda \approx \boxed{820\text{ lines/mm}}$$
Answer
$$f = 1/\Lambda \approx \boxed{820\text{ lines/mm}}$$
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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