---
title: "Minimum lines to resolve sodium lines"
subject: "Engineering Physics"
chapter: 4
key: "semester_1/engineering_physics:grating-minimum-lines-resolve-sodium"
source: https://www.bctengineer.com/q/semester_1/engineering_physics/grating-minimum-lines-resolve-sodium
site: BCT Engineering
---

# Minimum lines to resolve sodium lines

Engineering Physics · Chapter 4: Physical Optics

- Asked in: 2072 Kartik, 5 marks
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_1/engineering_physics/grating-minimum-lines-resolve-sodium
- Whole chapter: https://www.bctengineer.com/md/semester_1/engineering_physics/chapter-4

## Question

Calculate the minimum no of lines per cm in a 2.5 cm wide grating which will just resolve the sodium lines 5890 Å and 5896 Å in second order spectrum.

## Short answer

Resolving power R = λ/dλ = nN. With mean λ=5893 Å, dλ=6 Å, n=2, required N = 491.1 → 492 lines. Lines per cm = N/width = 491.1/2.5 = 196.4 → 197 lines/cm.

## Answer

Resolving power of a grating: R = λ/dλ = nN, where n is the order and N the total number of lines.

**Given:** λ₁ = 5890 Å, λ₂ = 5896 Å, n = 2, width = 2.5 cm.

1. Mean wavelength λ = (5890 + 5896)/2 = 5893 Å; dλ = 6 Å
2. Required resolving power:

```text
R = λ/dλ = 5893 / 6 = 982.2
```

3. Number of lines needed:

```text
N = R / n = 982.2 / 2 = 491.1  →  at least 492 lines
```

4. Lines per cm in the 2.5 cm wide grating:

```text
N / width = 491.1 / 2.5 = 196.4 lines/cm  →  at least 197 lines/cm
```

**Answer: the minimum is about 197 lines per cm (196.4 rounded up), i.e. about 492 lines across the 2.5 cm width.**

## Check yourself

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

1. The resolving power of a grating is:
   A. R = λ/dλ = nN
   B. R = N/n
   C. R = λ dλ
   D. R = n/N
   Answer: A. R = λ/dλ = nN, order times total number of lines. (Key point: Resolving power formula)

2. For 5890 Å and 5896 Å (mean 5893 Å), the required R is about:
   A. 982
   B. 5893
   C. 6
   D. 98
   Answer: A. R = 5893/6 ≈ 982. (Key point: Required resolving power)

3. Using R = nN with n = 2, the minimum total lines N is:
   A. 1964
   B. 982
   C. 491
   D. 245
   Answer: C. N = R/n = 982.2/2 = 491.1, so at least 492 lines. (Key point: Number of lines)

4. For a 2.5 cm wide grating, lines per cm is about:
   A. 1228
   B. 491
   C. 196
   D. 20
   Answer: C. 491.1/2.5 ≈ 196 lines/cm. (Key point: Lines per cm)

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