---
title: "Quarter wave plate thickness, quartz"
subject: "Engineering Physics"
chapter: 4
key: "semester_1/engineering_physics:quarter-wave-plate-quartz"
source: https://www.bctengineer.com/q/semester_1/engineering_physics/quarter-wave-plate-quartz
site: BCT Engineering
---

# Quarter wave plate thickness, quartz

Engineering Physics · Chapter 4: Physical Optics

- Asked in: 2073 Shrawan, 5 marks
- Video: [Quarter Wave & Half Wave Plates | Polarization | BTech, BSc & MSc Physics | CSIR NET GATE IIT JAM](https://www.youtube.com/watch?v=pOggQPaLQIA) — Dr Rajesh Kalia Physics 
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_1/engineering_physics/quarter-wave-plate-quartz
- Whole chapter: https://www.bctengineer.com/md/semester_1/engineering_physics/chapter-4

## Question

A quartz crystal has refractive indices 1.553 and 1.544. Calculate the thickness of a quarter wave plate for sodium light of wavelength 5890 Å.

## Short answer

Thickness t = λ / [4 (μe − μo)] = 5.890×10⁻⁷ m / (4 × 0.009) = 1.64 × 10⁻⁵ m (about 16.4 μm).

## Answer

A **quarter-wave plate** introduces a path difference of λ/4 between e and o rays, so (μ_e − μ_o) t = λ/4 (for the lowest-order plate). Quartz is a positive crystal: μ_e = 1.553 > μ_o = 1.544.

**Given:** λ = 5890 Å = 5.890×10⁻⁷ m.

1. Thickness:

```text
t = λ / [4 (μ_e − μ_o)]
  = 5.890×10⁻⁷ / [4 × (1.553 − 1.544)]
  = 5.890×10⁻⁷ / (4 × 0.009)
  = 5.890×10⁻⁷ / 0.036
  = 1.636×10⁻⁵ m
```

**Answer: t ≈ 1.64 × 10⁻⁵ m (about 16.4 μm).**

## Check yourself

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

1. What path difference does a quarter-wave plate introduce between e and o rays (lowest order)?
   A. λ/2
   B. λ/4
   C. λ
   D. 2λ
   Answer: B. A quarter-wave plate gives (μe − μo)t = λ/4, i.e. phase difference π/2. (Key point: QWP condition)

2. For quartz with indices 1.553 and 1.544, what is |μe − μo|?
   A. 0.009
   B. 0.09
   C. 3.097
   D. 0.0009
   Answer: A. 1.553 − 1.544 = 0.009. (Key point: Index difference)

3. Which thickness formula applies?
   A. t = λ / [2(μe − μo)]
   B. t = 4λ / (μe − μo)
   C. t = λ / [4(μe − μo)]
   D. t = λ(μe − μo)/4
   Answer: C. From (μe − μo)t = λ/4. (Key point: Formula)

4. The thickness for λ = 5890 Å is:
   A. About 1.64×10^-5 m
   B. About 1.64×10^-3 m
   C. About 6.5×10^-8 m
   D. About 2.6×10^-6 m
   Answer: A. t = 5.890×10^-7 / 0.036 = 1.636×10^-5 m, about 16.4 μm. (Key point: Numerical result)

## Related in Engineering Physics

- [Quarter wave plate thickness numerical](https://www.bctengineer.com/question.md?key=semester_1%2Fengineering_physics%3Aquarter-wave-plate-thickness) — Engineering Physics, chapter 4
- [Polarization; quarter and half wave plates](https://www.bctengineer.com/question.md?key=semester_1%2Fengineering_physics%3Apolarization-importance-quarter-half-wave) — Engineering Physics, chapter 4
- [Thickness of doubly refracting plate](https://www.bctengineer.com/question.md?key=semester_1%2Fengineering_physics%3Adoubly-refracting-plate-thickness) — Engineering Physics, chapter 4