---
title: "Second order maxima in single slit"
subject: "Engineering Physics"
chapter: 4
key: "semester_1/engineering_physics:single-slit-second-order-maxima-ratio"
source: https://www.bctengineer.com/q/semester_1/engineering_physics/single-slit-second-order-maxima-ratio
site: BCT Engineering
---

# Second order maxima in single slit

Engineering Physics · Chapter 4: Physical Optics

- Asked in: 2067 Asar, 5 marks
- Video: [Maxima in Single slit diffraction | Wave optics | Physics | Khan Academy](https://www.youtube.com/watch?v=u6aGVO2smyw) — Khan Academy India - English
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_1/engineering_physics/single-slit-second-order-maxima-ratio
- Whole chapter: https://www.bctengineer.com/md/semester_1/engineering_physics/chapter-4

## Question

Show that the intensity of second order maxima of Fraunhofer's single slit diffraction is 2/(5π) times the intensity of central maxima.

## Short answer

The second-order maximum occurs at α ≈ 5π/2, where sinα/α = 2/(5π). This is the amplitude ratio relative to the central maximum; the intensity ratio is (2/(5π))² = 4/(25π²) ≈ 0.0162.

## Answer

#### Intensity in single-slit diffraction
For a slit of width *a*, the resultant amplitude at angle θ is R = R₀ (sin α)/α, and the intensity is

I = I₀ (sin α / α)², where α = (π a sin θ) / λ.

The central maximum is at α = 0, where sin α / α = 1, so I = I₀.

#### Position of the secondary maxima
Secondary maxima occur where dI/dα = 0, i.e. tan α = α. Solving graphically gives α ≈ ±1.43π, ±2.46π, ... A very good approximation, which is the usual textbook one, is that the maxima lie midway between the minima:

α ≈ (2n + 1)π / 2, n = 1, 2, 3, ...

n = 1 is the first secondary (first order) maximum with α = 3π/2; **n = 2 is the second-order maximum with α = 5π/2**.

#### Second-order maximum
```text
sin α  = sin (5π/2) = 1
sin α / α = 1 / (5π/2) = 2 / (5π)       (amplitude ratio)
```
So the **amplitude** of the second-order maximum is **2/(5π)** times the central amplitude (≈ 0.127).

The **intensity** is the square of this:

I₂ / I₀ = (2 / 5π)² = 4 / (25π²) ≈ 0.0162 (about 1/62).

**Remark:** the figure 2/(5π) is the ratio R₂/R₀ of the amplitudes (the ratio sin α / α). The ratio of the intensities is its square, 4/(25π²). The first-order maximum similarly gives 2/(3π) for amplitude and 4/(9π²) ≈ 0.045 for intensity.

```text
 I/I0
 1.0 |        *
     |       * *
     |      *   *
 0.045|  *  *     *  *   <- 1st secondary
 0.016|*                  <- 2nd secondary
     +-------------------> sin(theta)
```

## Check yourself

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

1. In single-slit diffraction, how is the intensity related to I0?
   A. I = I0 (sin α / α)
   B. I = I0 (sin α / α)^2
   C. I = I0 (α / sin α)^2
   D. I = I0 sin^2 α
   Answer: B. Amplitude is R0 sin α/α and intensity is its square. (Key point: Intensity formula)

2. Which value of α is used for the second-order secondary maximum?
   A. 3π/2
   B. 5π/2
   C. 2π
   D. π
   Answer: B. Maxima are taken at α = (2n+1)π/2; n = 2 gives 5π/2. (Key point: Second-order maximum)

3. What does the quantity 2/(5π) actually represent?
   A. Intensity ratio I2/I0
   B. Amplitude ratio R2/R0
   C. Ratio of slit widths
   D. Angular position of the maximum
   Answer: B. sin α / α = 2/(5π) is the amplitude ratio; the intensity ratio is its square. (Key point: Meaning of 2/(5π))

4. What is the true intensity ratio I2/I0 for the second secondary maximum?
   A. 2/(5π)
   B. 4/(25π^2) ≈ 0.0162
   C. 2/(3π)
   D. 1/(5π)
   Answer: B. Squaring the amplitude ratio gives 4/(25π^2) ≈ 1/62. (Key point: Intensity ratio)

## Close variants set in other exams

- [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×

## 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

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