---
title: "Separation of second minima in single slit"
subject: "Engineering Physics"
chapter: 4
key: "semester_1/engineering_physics:fraunhofer-slit-separation-second-minima"
source: https://www.bctengineer.com/q/semester_1/engineering_physics/fraunhofer-slit-separation-second-minima
site: BCT Engineering
---

# Separation of second minima in single slit

Engineering Physics · Chapter 4: Physical Optics

- Asked in: 2072 Kartik, 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/fraunhofer-slit-separation-second-minima
- Whole chapter: https://www.bctengineer.com/md/semester_1/engineering_physics/chapter-4

## Question

In a Fraunhofer single slit diffraction, a convex lens of focal length 20 cm is placed just after a slit of width 0.6 mm. If a plane wave of wavelength 6000 Å falls on the slit normally, calculate the separation between the second minima on either side of the central maximum.

## Short answer

Minima satisfy a sinθ = mλ. Distance from center is x ≈ f mλ/a, so x₂ = 0.4 mm for m = 2. Separation between the two second minima is 2x₂ = 0.8 mm.

## Answer

In single-slit Fraunhofer diffraction, minima occur at a sin θ = mλ. With a lens of focal length *f*, a minimum is at distance x = f tan θ ≈ f sin θ from the centre.

**Given:** f = 20 cm = 0.20 m, a = 0.6 mm = 6×10⁻⁴ m, λ = 6000 Å = 6×10⁻⁷ m, m = 2.

1. Angular position of second minimum:

```text
sin θ₂ = 2λ / a = 2 × 6×10⁻⁷ / 6×10⁻⁴ = 2×10⁻³  (θ small)
```

2. Distance from centre:

```text
x₂ = f sin θ₂ = 0.20 × 2×10⁻³ = 4×10⁻⁴ m = 0.4 mm
```

3. The two second minima lie on either side of the central maximum, so their separation is

```text
2 x₂ = 2 × 0.4 mm = 0.8 mm
```

**Answer: separation between the two second minima = 8 × 10⁻⁴ m (0.8 mm).**

## Check yourself

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

1. In a single slit, minima occur when:
   A. a sin θ = mλ
   B. a sin θ = (m + 1/2)λ
   C. a cos θ = mλ
   D. a sin θ = mλ/2
   Answer: A. Single slit minima: a sin θ = mλ, m = 1, 2, ... (Key point: Minima condition)

2. With a lens of focal length f, the distance of a minimum from the centre is:
   A. x = f / sin θ
   B. x = f sin θ (≈ f tan θ)
   C. x = λ f
   D. x = a f
   Answer: B. The lens focuses angle θ at x = f tan θ ≈ f sin θ. (Key point: Position on screen)

3. What is x2 for f = 20 cm, a = 0.6 mm, λ = 6000 Å?
   A. 0.4 mm
   B. 4 mm
   C. 0.04 mm
   D. 0.2 mm
   Answer: A. sin θ2 = 2×10^-3, so x2 = 0.20 × 2×10^-3 = 4×10^-4 m = 0.4 mm. (Key point: Second minimum position)

4. The separation between the two second minima is:
   A. 0.4 mm
   B. 0.8 mm
   C. 0.2 mm
   D. 1.6 mm
   Answer: B. They lie on opposite sides, so the separation is 2x2 = 0.8 mm. (Key point: Separation)

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