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

# Quarter wave plate thickness numerical

Engineering Physics · Chapter 4: Physical Optics

- Asked in: 2067 Asar, 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-thickness
- Whole chapter: https://www.bctengineer.com/md/semester_1/engineering_physics/chapter-4

## Question

Calculate the thickness of quarter wave plate for light of wavelength 5893 Å. Given refractive indices of ordinary and extraordinary ray 1.544 and 1.553 respectively.

## Short answer

For a quarter-wave plate, t = λ/[4(μ_e−μ_o)]. With λ=5893 Å, μ_e−μ_o=0.009, t = 1.64×10⁻⁵ m (16.4 µm).

## Answer

A quarter-wave plate is a doubly refracting crystal plate (such as quartz or calcite) cut with its optic axis parallel to its faces. It introduces a path difference of λ/4 (a phase difference of π/2) between the ordinary and extraordinary rays.

**Condition:** (μ_e − μ_o) t = λ/4 (for a positive crystal such as quartz, μ_e > μ_o), so

t = λ / [4 (μ_e − μ_o)]

**Given:** λ = 5893 Å = 5.893 × 10⁻⁷ m, μ_o = 1.544, μ_e = 1.553.

```text
μ_e − μ_o = 1.553 − 1.544 = 0.009

t = 5.893×10⁻⁷ / (4 × 0.009)
  = 5.893×10⁻⁷ / 0.036
  = 1.637×10⁻⁵ m
```
**Answer: t ≈ 1.64 × 10⁻⁵ m = 16.4 µm (1.64 × 10⁻³ cm).**

This is the minimum thickness; any odd multiple, (2m + 1)λ / 4(μ_e − μ_o), also works.

## Check yourself

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

1. What path difference must a quarter-wave plate introduce between the o and e rays?
   A. λ
   B. λ/2
   C. λ/4
   D. λ/8
   Answer: C. A quarter-wave plate gives λ/4, i.e. a phase difference of π/2. (Key point: Path difference condition)

2. Which formula gives the minimum thickness?
   A. t = λ / [4(μ_e + μ_o)]
   B. t = λ / [4(μ_e − μ_o)]
   C. t = 4λ / (μ_e − μ_o)
   D. t = λ(μ_e − μ_o)/4
   Answer: B. (μ_e − μ_o) t = λ/4, so t = λ/[4(μ_e − μ_o)]. (Key point: Thickness formula)

3. What is t for λ = 5893 Å, μ_o = 1.544, μ_e = 1.553?
   A. 1.64 × 10^-5 m
   B. 1.64 × 10^-7 m
   C. 6.5 × 10^-4 m
   D. 1.64 × 10^-3 m
   Answer: A. t = 5.893×10^-7 / (4 × 0.009) = 1.64 × 10^-5 m, about 16.4 µm. (Key point: Numerical result)

## Related in Engineering Physics

- [Quarter wave plate thickness, quartz](https://www.bctengineer.com/question.md?key=semester_1%2Fengineering_physics%3Aquarter-wave-plate-quartz) — 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
- [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