---
title: "Index difference for circular polarization"
subject: "Engineering Physics"
chapter: 4
key: "semester_1/engineering_physics:qwp-refractive-index-difference"
source: https://www.bctengineer.com/q/semester_1/engineering_physics/qwp-refractive-index-difference
site: BCT Engineering
---

# Index difference for circular polarization

Engineering Physics · Chapter 4: Physical Optics

- Asked in: 2070 Asar, 5 marks
- Video: [Polarizations and Waveplates](https://www.youtube.com/watch?v=Pm5aKYIOccc) — Reumi's world
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_1/engineering_physics/qwp-refractive-index-difference
- Whole chapter: https://www.bctengineer.com/md/semester_1/engineering_physics/chapter-4

## Question

A beam of plane polarized light is converted into circularly polarized light by passing it through a crystal slice of thickness 3×10⁻⁵ m. Calculate the difference in the refractive indices of the two rays inside the crystal. Wavelength of light is 600nm.

## Short answer

For a quarter-wave plate, path difference (μ_e−μ_o)t = λ/4. Thus |μ_e−μ_o| = λ/(4t) = 600×10⁻⁹/(4×3×10⁻⁵) = 0.005.

## Answer

Plane polarized light becomes circularly polarized after passing through a crystal slice (quarter-wave plate) when the path difference between the e and o rays is λ/4, i.e. (μ_e − μ_o) t = λ/4 (lowest order, with vibration at 45° to the axis).

**Given:** t = 3×10⁻⁵ m, λ = 600 nm = 6×10⁻⁷ m.

1. Condition: |μ_e − μ_o| t = λ/4
2. Solve:

```text
|μ_e − μ_o| = λ / (4t)
             = 6×10⁻⁷ / (4 × 3×10⁻⁵)
             = 6×10⁻⁷ / 1.2×10⁻⁴
             = 5×10⁻³
```

**Answer: difference in refractive indices = 0.005.**

## Check yourself

The answer's key points as multiple-choice questions, with the correct option and why.

1. Plane polarized light becomes circularly polarized after a crystal slice when the path difference between e and o rays is:
   A. λ/2
   B. λ/4
   C. λ
   D. λ/8
   Answer: B. A quarter-wave plate (lowest order) gives λ/4 with vibration at 45° to the axis. (Key point: Condition for circular light)

2. Which equation is used?
   A. |μe − μo| t = λ/4
   B. |μe − μo| = λ t / 4
   C. |μe + μo| t = λ/4
   D. |μe − μo| = 4λt
   Answer: A. Path difference (μe − μo)t = λ/4. (Key point: Equation)

3. For t = 3×10^-5 m and λ = 600 nm, |μe − μo| is:
   A. 0.005
   B. 0.05
   C. 0.0005
   D. 0.02
   Answer: A. 6×10^-7 / (4×3×10^-5) = 5×10^-3. (Key point: Result)

## Related in Engineering Physics

- [Production of plane, circular, elliptical light](https://www.bctengineer.com/question.md?key=semester_1%2Fengineering_physics%3Apolarized-light-production-relations) — Engineering Physics, chapter 4, asked 3×
- [Distinguishing plane, circular, elliptical light](https://www.bctengineer.com/question.md?key=semester_1%2Fengineering_physics%3Adistinguish-polarized-nicol-wave-plate) — Engineering Physics, chapter 4
- [Quarter wave plate thickness, quartz](https://www.bctengineer.com/question.md?key=semester_1%2Fengineering_physics%3Aquarter-wave-plate-quartz) — Engineering Physics, chapter 4
- [Double refraction, linear and circular light](https://www.bctengineer.com/question.md?key=semester_1%2Fengineering_physics%3Adouble-refraction-produce-polarized-light) — Engineering Physics, chapter 4
- [Newton's rings, refractive index of liquid](https://www.bctengineer.com/question.md?key=semester_1%2Fengineering_physics%3Anewton-rings-liquid-refractive-index) — Engineering Physics, chapter 4, asked 2×
- [Refractive index of liquid film by interference](https://www.bctengineer.com/question.md?key=semester_1%2Fengineering_physics%3Acoherent-sources-film-refractive-index-method) — Engineering Physics, chapter 4

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