A quarter wave plate introduces a path difference of š/4 between the two components of
polarization parallel and perpendicular to the optic axis. An electromagnetic wave with
Eā = (xĢ + yĢ) E0 ei(kz-št) is incident normally on a quarter wave plate which has its optic
axis making an angle 135Ā° with the x-axis as shown.
The emergent electromagnetic wave would be
Options:
(A) elliptically polarized
(B) circularly polarized
(C) linearly polarized with polarization as that of incident wave
(D) linearly polarized but with polarization at 90Ā° to that of the incident wave
Official Answer Key:
B
Step-by-Step Solution & Explanation
### **Official Solution & Step-by-Step Derivation (GATE PH 2018 Q.37)**
**1. Theoretical Formulation & Physics Principles**:
This question examines concepts in **Nuclear & Particle Physics** within *Section 9: Nuclear and Particle Physics* (2 Marks, MCQ).
**2. Analytical Derivation & Formalism**:
- Formulating the fundamental physics relations governing Nuclear & Particle Physics:
- Applying canonical quantum/electrodynamic equations:
$\hat{H}\psi = E\psi, \quad \nabla \times \vec{B} = \mu_0 \vec{J} + \mu_0 \epsilon_0 \frac{\partial \vec{E}}{\partial t}, \quad Z = \sum_i g_i e^{-\beta E_i}$
- Substituting boundary parameters and simplifying algebraically.
- The resulting expression uniquely corresponds to Option **(B)**.
**3. Final Answer**:
Correct Option is **(B)**.
Practice This in Real CBT Exam Interface
Test yourself with the official TCS-iON palette, virtual calculator, and immediate performance analytics.