---
title: "Single slit diffraction intensity distribution"
subject: "Engineering Physics"
chapter: 4
key: "semester_1/engineering_physics:single-slit-intensity-distribution"
source: https://www.bctengineer.com/q/semester_1/engineering_physics/single-slit-intensity-distribution
site: BCT Engineering
---

# Single slit diffraction intensity distribution

Engineering Physics · Chapter 4: Physical Optics

- Asked in: 2072 Chaitra, 5 marks; 2070 Chaitra, 5 marks; 2068 Shrawan, 5 marks (asked 3 times)
- Video: [Intensity distribution in the diffraction pattern of a single slit [Fraunhofer Diffraction - Part 2]](https://www.youtube.com/watch?v=s_PXirMXTEg) — Time for LEARNING PLUS
- 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-intensity-distribution
- Whole chapter: https://www.bctengineer.com/md/semester_1/engineering_physics/chapter-4

## Question

What is diffraction of light? Discuss the intensity distribution of Fraunhofer diffraction at a single slit, with special reference to the pattern I = I0(sin α/α)². Show that the relative intensities of the successive maxima are 1 : 4/(9π²) : 4/(25π²) ...

## Short answer

Diffraction is bending of light around obstacles. Single-slit intensity I = I₀(sinα/α)² with α = πa sinθ/λ. Minima at a sinθ = mλ; secondary maxima near α = (2m+1)π/2, with relative intensities 1 : 4/(9π²) : 4/(25π²) i.e. 1 : 0.045 : 0.016.

## Answer

**Diffraction** is the bending of light waves around the edges of obstacles/apertures of size comparable with λ. In **Fraunhofer** diffraction the source and screen are effectively at infinite distance (plane wavefronts, using lenses).

#### Intensity distribution at a single slit
```text
 slit width a                 screen
   |  \  θ  |                 I
 --|-- \    |                /\
   |    \ ->|--------  ... _/  \_  ...
```
1. A parallel beam falls normally on a slit of width *a*. Divide the slit into *N* narrow strips; each sends a secondary wavelet of equal amplitude.
2. Path difference between the extreme rays at angle θ is a sin θ, so the total phase difference is 2α = (2π/λ) a sin θ, i.e. **α = π a sin θ / λ**.
3. The vector (phasor) sum of N small equal vectors with total phase 2α forms an arc of a circle. With arc length = A₀ (the amplitude at θ = 0) and radius r: R = 2r sin α, A₀ = 2rα.
4. So the resultant amplitude: **R = A₀ (sin α)/α**.
5. Intensity I = R²:

```text
I = I₀ (sin α / α)² ,    α = π a sin θ / λ
```

#### Minima and maxima
- **Central maximum** at α → 0: sinα/α → 1, so I = I₀.
- **Minima**: sin α = 0, α = ±mπ (m = 1, 2 ...), i.e. **a sin θ = mλ**.
- **Secondary maxima**: dI/dα = 0 gives tan α = α, α ≈ ±3π/2, ±5π/2 ... (approximately, midway between minima).

#### Relative intensities
At α = (2m+1)π/2, sin²α = 1, so I_m = I₀ / α² = 4 I₀ / [(2m+1)² π²].

```text
I₀ : I₁ : I₂ : ... = 1 : 4/(9π²) : 4/(25π²) : ...
                    = 1 : 0.045 : 0.016 : ...
```

So the first secondary maximum is only ~4.5% and the second ~1.6% of the central maximum: intensity falls rapidly, and most light lies in the central maximum, whose width is 2λ/a (angular).

## Check yourself

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

1. In I = I0(sin α/α)², α equals
   A. π a sin θ / λ
   B. 2π a sin θ / λ
   C. a sin θ / λ
   D. π λ / (a sin θ)
   Answer: A. The total phase difference is 2α = (2π/λ)a sin θ, so α = πa sin θ/λ. (Key point: Definition of alpha)

2. The minima of the single-slit pattern occur at
   A. a sin θ = (m + ½)λ
   B. a sin θ = mλ
   C. a sin θ = 0 for all m
   D. a sin θ = mλ/2
   Answer: B. sin α = 0 gives α = mπ, i.e. a sin θ = mλ (m = 1, 2, ...). (Key point: Minima)

3. The secondary maxima are given approximately by
   A. α = mπ
   B. tan α = α, α ≈ (2m+1)π/2
   C. α = 0
   D. sin α = α
   Answer: B. dI/dα = 0 gives tan α = α with roots near 3π/2, 5π/2, ... midway between minima. (Key point: Secondary maxima)

4. The relative intensities of successive maxima are
   A. 1 : 1/9 : 1/25
   B. 1 : 4/(9π²) : 4/(25π²)
   C. 1 : 1/3 : 1/5
   D. 1 : 2/π : 2/(3π)
   Answer: B. At α = (2m+1)π/2, sin²α = 1 so I = I0/α² = 4I0/((2m+1)²π²). (Key point: Relative intensities)

## Related in Engineering Physics

- [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×
- [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×
- [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×
- [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
- [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
- [Intensity distribution in interference](https://www.bctengineer.com/question.md?key=semester_1%2Fengineering_physics%3Ainterference-intensity-resultant) — Engineering Physics, chapter 4, asked 2×

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