---
title: "First subsidiary maxima intensity percentage"
subject: "Engineering Physics"
chapter: 4
key: "semester_1/engineering_physics:fraunhofer-first-maxima-percent"
source: https://www.bctengineer.com/q/semester_1/engineering_physics/fraunhofer-first-maxima-percent
site: BCT Engineering
---

# First subsidiary maxima intensity percentage

Engineering Physics · Chapter 4: Physical Optics

- Asked in: 2073 Shrawan, 5 marks; 2069 Chaitra, 5 marks (asked 2 times)
- Video: [Fraunhofer Single Slit Diffraction, Resultant Amplitude & Intensity AKTU BTech 1st Year Eng Physics](https://www.youtube.com/watch?v=Gj57RkylsHQ) — EduGlue AKTU
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_1/engineering_physics/fraunhofer-first-maxima-percent
- Whole chapter: https://www.bctengineer.com/md/semester_1/engineering_physics/chapter-4

## Question

Prove that the intensity of the first (subsidiary) maxima is 4.54% (about 4.5%) of the central (principal) maxima in Fraunhofer's single slit diffraction.

## Short answer

In single-slit diffraction I=I₀(sinα/α)². The first subsidiary maximum near α=3π/2 yields I₁=I₀/(3π/2)²=4I₀/(9π²)≈0.045 I₀, about 4.5% of the central maximum.

## Answer

In single-slit Fraunhofer diffraction, I = I₀ (sin α / α)², where α = π a sin θ/λ.

#### Positions of maxima
1. For a maximum, dI/dα = 0:
   dI/dα = I₀ · 2 (sin α/α) · (α cos α − sin α)/α² = 0
2. Either sin α = 0 (minima) or α cos α − sin α = 0, i.e. **tan α = α**.
3. The roots (graphically from the intersections of y = tan α and y = α) are close to α ≈ ±3π/2, ±5π/2 ... midway between the minima at ±π, ±2π, ... Using the textbook approximation, the *m*th secondary maximum is at α ≈ (2m+1)π/2.

```text
 y
 |   y = α   /  /   /
 |          /  /   /     tan α curves
 |-----π---/--3π/2--2π--5π/2----> α
```

#### Intensity of the first secondary maximum
Put α = 3π/2: sin α = −1, so

```text
I₁ = I₀ (1/α)² = I₀ / (3π/2)² = 4 I₀ / (9π²)
   = 4 / (9 × 9.8696) I₀ = 0.045 I₀
```

So I₁/I₀ ≈ 4.5 %. (Solving tan α = α exactly gives α = 4.493 and 4.7 %; the commonly quoted textbook value is about 4.5 %.)

**Hence the first subsidiary maximum has only about 4.5 % of the intensity of the principal (central) maximum.**

## Check yourself

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

1. Setting dI/dα = 0 for I = I0(sin α/α)² gives the maxima condition
   A. sin α = α
   B. tan α = α
   C. cos α = α
   D. tan α = 0
   Answer: B. α cos α - sin α = 0, i.e. tan α = α, for the secondary maxima. (Key point: Condition for maxima)

2. The first secondary maximum is taken (approximately) at
   A. α = π
   B. α = π/2
   C. α = 2π
   D. α = 3π/2
   Answer: D. The maxima lie midway between minima at π and 2π, i.e. near 3π/2. (Key point: Position of first secondary maximum)

3. At α = 3π/2, I/I0 equals
   A. 4/(9π²)
   B. 1/(3π)
   C. 2/(3π)
   D. 9π²/4
   Answer: A. sin²α = 1, so I = I0/α² = I0/(3π/2)² = 4I0/(9π²). (Key point: Intensity there)

4. This corresponds to about what percentage of the central maximum?
   A. 45%
   B. 0.45%
   C. 4.5%
   D. 14%
   Answer: C. 4/(9 × 9.87) ≈ 0.045, i.e. about 4.5%. (Key point: Numerical value)

## Related in Engineering Physics

- [Single slit diffraction intensity distribution](https://www.bctengineer.com/question.md?key=semester_1%2Fengineering_physics%3Asingle-slit-intensity-distribution) — Engineering Physics, chapter 4, asked 3×
- [Second order maxima in single slit](https://www.bctengineer.com/question.md?key=semester_1%2Fengineering_physics%3Asingle-slit-second-order-maxima-ratio) — Engineering Physics, chapter 4, asked 2×
- [Second maxima intensity 1.62 percent](https://www.bctengineer.com/question.md?key=semester_1%2Fengineering_physics%3Asecond-maxima-intensity-percent) — Engineering Physics, chapter 4, asked 2×
- [Separation of second minima in single slit](https://www.bctengineer.com/question.md?key=semester_1%2Fengineering_physics%3Afraunhofer-slit-separation-second-minima) — Engineering Physics, chapter 4
- [Single slit intensity at given angle](https://www.bctengineer.com/question.md?key=semester_1%2Fengineering_physics%3Asingle-slit-intensity-at-angle) — Engineering Physics, chapter 4