Problem Statement
Unpolarized light falls on a glass surface ($n = 1.5$) at Brewster’s angle. Find the degree of polarization of the reflected beam.
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: At Brewster’s angle the reflected beam contains only the $s$-component (perpendicular to plane of incidence). The degree of polarization is:
Step 2 — Apply the relevant physical law or equation: $$P = \frac{I_{\max}-I_{\min}}{I_{\max}+I_{\min}} = \frac{I_s-0}{I_s+0} = \boxed{1.0}$$
Step 3 — Solve algebraically for the unknown: The reflected beam is 100% linearly polarized.
Worked Calculation
$$P = \frac{I_{\max}-I_{\min}}{I_{\max}+I_{\min}} = \frac{I_s-0}{I_s+0} = \boxed{1.0}$$
$$\text{Numerical result} = \text{given expression substituted with values}$$
$$\boxed{P = \frac{I_{\max}-I_{\min}}{I_{\max}+I_{\min}} = \frac{I_s-0}{I_s+0} = \boxed{1.0}}$$
At Brewster’s angle the reflected beam contains only the $s$-component (perpendicular to plane of incidence). The degree of polarization is:
$$P = \frac{I_{\max}-I_{\min}}{I_{\max}+I_{\min}} = \frac{I_s-0}{I_s+0} = \boxed{1.0}$$
The reflected beam is 100% linearly polarized.
Answer
$$P = \frac{I_{\max}-I_{\min}}{I_{\max}+I_{\min}} = \frac{I_s-0}{I_s+0} = \boxed{1.0}$$
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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