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

# Single slit intensity at given angle

Engineering Physics · Chapter 4: Physical Optics

- Asked in: 2069 Asar, 5 marks
- Video: [Single Slit Diffraction - Physics Problems](https://www.youtube.com/watch?v=9hCrhllI0ck) — The Organic Chemistry Tutor
- 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-at-angle
- Whole chapter: https://www.bctengineer.com/md/semester_1/engineering_physics/chapter-4

## Question

Light of wavelength 600nm is incident normally on a slit of width 0.1mm. Calculate the intensity at θ=0.2°.

## Short answer

α = π a sinθ/λ = π×10⁻⁴×sin(0.2°)/6×10⁻⁷ ≈ 1.828 rad. I = I₀ (sinα/α)² ≈ 0.28 I₀.

## Answer

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

**Given:** λ = 600 nm = 6×10⁻⁷ m, a = 0.1 mm = 1×10⁻⁴ m, θ = 0.2°.

1. Angle in radians: θ = 0.2 × π/180 = 3.491×10⁻³ rad, so sin θ ≈ 3.491×10⁻³
2. Compute α:

```text
α = π a sin θ / λ
  = π × 1×10⁻⁴ × 3.491×10⁻³ / 6×10⁻⁷
  = 1.828 rad
```

3. sin α = sin(1.828 rad) = 0.9672, so sin α / α = 0.9672/1.828 = 0.5292
4. Intensity:

```text
I = I₀ (0.5292)² = 0.280 I₀
```

**Answer: I ≈ 0.28 I₀, i.e. about 28 % of the central maximum intensity.**

## Check yourself

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

1. What is α in the single slit intensity formula?
   A. α = π a sin θ / λ
   B. α = a sin θ / λ
   C. α = 2π λ / (a sin θ)
   D. α = π λ sin θ / a
   Answer: A. α = π a sin θ/λ is half the phase difference across the slit. (Key point: Formula)

2. For a = 0.1 mm, λ = 600 nm, θ = 0.2°, α is about:
   A. 1.83 rad
   B. 0.18 rad
   C. 18.3 rad
   D. 0.58 rad
   Answer: A. α = π×10^-4×3.491×10^-3/6×10^-7 = 1.828 rad. (Key point: Calculating α)

3. The intensity at this angle is about:
   A. 0.28 I0
   B. 0.53 I0
   C. 0.97 I0
   D. 0.05 I0
   Answer: A. (sin α/α)^2 = (0.9672/1.828)^2 = 0.28. (Key point: Intensity)

## 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
- [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×
- [Double slit fringe separation numerical](https://www.bctengineer.com/question.md?key=semester_1%2Fengineering_physics%3Ayoung-double-slit-fringe-separation) — Engineering Physics, chapter 4
- [Fringe width of air wedge numerical](https://www.bctengineer.com/question.md?key=semester_1%2Fengineering_physics%3Awedge-fringe-width-wire-numerical) — Engineering Physics, chapter 4
- [Wedge angle from fringe spacing](https://www.bctengineer.com/question.md?key=semester_1%2Fengineering_physics%3Awedge-angle-fringe-spacing) — Engineering Physics, chapter 4

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