Problem 5.161 — Circular Polarization Detection

Problem Statement

A beam is suspected to be circularly polarized. Describe how to verify this using a quarter-wave plate and a polarizer, and what result confirms circular polarization.

Given Information

• See problem statement for all given quantities.

Physical Concepts & Formulas

Circular motion requires a centripetal force directed toward the centre, providing the centripetal acceleration $a_c = v^2/r = \omega^2 r$. This force is not a new type of force — it is always the resultant of real forces (tension, normal force, friction, gravity) directed inward. At the minimum speed for maintaining contact, the normal force drops to zero.

• $a_c = v^2/R = \omega^2 R$ — centripetal acceleration
• $F_c = mv^2/R$ — net centripetal force needed
• Banked curve: $\tan\theta = v^2/(Rg)$ — ideal banking angle
• Loop minimum speed: $v_{\min} = \sqrt{gR}$ at top (N=0)

Step-by-Step Solution

Step 1 — Identify given quantities and set up the problem: Procedure:

Step 2 — Apply the relevant physical law or equation:

1. First, rotate a linear polarizer alone: if intensity is constant for all angles, the beam could be circularly polarized or unpolarized.
2. Insert a quarter-wave plate before the polarizer. If the resulting beam through the polarizer shows a maximum and minimum (varies as $\cos^2\theta$), the original beam was circularly polarized (the QWP converts it to linear).
3. If still constant, the original beam was unpolarized.

Step 3 — Solve algebraically for the unknown: $$\boxed{\text{After QWP + polarizer: constant intensity} \Rightarrow \text{unpolarized; } I \propto \cos^2\theta \Rightarrow \text{circularly polarized}}$$

Worked Calculation

$$\boxed{\text{After QWP + polarizer: constant intensity} \Rightarrow \text{unpolarized; } I \propto \cos^2\theta \Rightarrow \text{circularly polarized}}$$

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

$$\boxed{\boxed{\text{After QWP + polarizer: constant intensity} \Rightarrow \text{unpolarized; } I \propto \cos^2\theta \Rightarrow \text{circularly polarized}}}$$

Procedure:

1. First, rotate a linear polarizer alone: if intensity is constant for all angles, the beam could be circularly polarized or unpolarized.
2. Insert a quarter-wave plate before the polarizer. If the resulting beam through the polarizer shows a maximum and minimum (varies as $\cos^2\theta$), the original beam was circularly polarized (the QWP converts it to linear).
3. If still constant, the original beam was unpolarized.

$$\boxed{\text{After QWP + polarizer: constant intensity} \Rightarrow \text{unpolarized; } I \propto \cos^2\theta \Rightarrow \text{circularly polarized}}$$

Answer

$$\boxed{\text{After QWP + polarizer: constant intensity} \Rightarrow \text{unpolarized; } I \propto \cos^2\theta \Rightarrow \text{circularly polarized}}$$

Physical Interpretation

The centripetal force is not a ‘new’ force but the net inward resultant of real forces. If that resultant falls below $mv^2/r$, the object cannot maintain circular motion and will fly outward — this is the critical condition for minimum speed problems.