---
title: "Second maxima intensity 1.62 percent"
subject: "Engineering Physics"
chapter: 4
key: "semester_1/engineering_physics:second-maxima-intensity-percent"
source: https://www.bctengineer.com/q/semester_1/engineering_physics/second-maxima-intensity-percent
site: BCT Engineering
---

# Second maxima intensity 1.62 percent

Engineering Physics · Chapter 4: Physical Optics

- Asked in: 2070 Asar, 5 marks
- 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/second-maxima-intensity-percent
- Whole chapter: https://www.bctengineer.com/md/semester_1/engineering_physics/chapter-4

## Question

Show that the intensity of second primary maxima is 1.62% of central maxima in Fraunhofer's single slit diffraction.

## Short answer

In single-slit diffraction, I = I₀(sinα/α)². The second secondary maximum occurs at α ≈ 5π/2 where sinα=1, giving I₂/I₀ = 4/(25π²) ≈ 0.0162 or 1.62%.

## Answer

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

1. Maxima: dI/dα = 0 gives α cos α − sin α = 0, i.e. tan α = α.
2. The roots lie close to the odd multiples of π/2 (midway between the minima α = mπ): α ≈ ±3π/2 for the first, ±**5π/2** for the second secondary maximum.

```text
 I/I0
 1 |        /\
   |       /  \
.045|  _   /    \   _
.016|_/ \_/      \_/ \_
   +--------------------> α
   -5π/2 -3π/2  0  3π/2 5π/2
```

3. At α = 5π/2, sin α = 1, so

```text
I₂ = I₀ (1/α)² = I₀ / (5π/2)² = 4 I₀ / (25 π²)
   = 4 / (25 × 9.8696) I₀
   = 0.0162 I₀
```

4. As a percentage: I₂/I₀ ≈ 1.62 %.

(The first secondary maximum, at 3π/2, similarly gives 4/(9π²) ≈ 4.5 %.)

**Answer: the second secondary maximum has about 1.62 % of the central maximum intensity.**

## Check yourself

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

1. In single slit diffraction, I equals:
   A. I0 (sin α / α)^2
   B. I0 (α / sin α)^2
   C. I0 sin α
   D. I0 cos^2 α
   Answer: A. I = I0 (sin α/α)^2 with α = π a sin θ/λ. (Key point: Intensity formula)

2. The secondary maxima satisfy:
   A. sin α = 0
   B. tan α = α
   C. cos α = 0 exactly
   D. α = mπ
   Answer: B. dI/dα = 0 gives α cos α = sin α, i.e. tan α = α, with roots near odd multiples of π/2. (Key point: Locating maxima)

3. The second secondary maximum is approximately at α =
   A. 3π/2
   B. 5π/2
   C. 2π
   D. π
   Answer: B. The first is near 3π/2 and the second near 5π/2. (Key point: Second maximum position)

4. At α = 5π/2, I/I0 =
   A. 4/(25π^2) ≈ 0.0162
   B. 4/(9π^2) ≈ 0.045
   C. 1/(5π) ≈ 0.064
   D. 2/(5π^2)
   Answer: A. sin α = 1, so I = I0/(5π/2)^2 = 4I0/(25π^2) ≈ 1.62%. (Key point: Calculation)

## Close variants set in other exams

- [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×

## 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×
- [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
- [First subsidiary maxima intensity percentage](https://www.bctengineer.com/question.md?key=semester_1%2Fengineering_physics%3Afraunhofer-first-maxima-percent) — Engineering Physics, chapter 4, asked 2×
- [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