---
title: "Grating lines per cm, superposed lines"
subject: "Engineering Physics"
chapter: 4
key: "semester_1/engineering_physics:grating-lines-per-cm-superposed-orders"
source: https://www.bctengineer.com/q/semester_1/engineering_physics/grating-lines-per-cm-superposed-orders
site: BCT Engineering
---

# Grating lines per cm, superposed lines

Engineering Physics · Chapter 4: Physical Optics

- Asked in: 2074 Asoj, 5 marks; 2067 Asoj, 5 marks (asked 2 times)
- Video: [Diffraction Grating Problems - Physics](https://www.youtube.com/watch?v=gf7j2fumz70) — 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/grating-lines-per-cm-superposed-orders
- Whole chapter: https://www.bctengineer.com/md/semester_1/engineering_physics/chapter-4

## Question

A diffraction grating used at normal incidence gives a line (540 nm) in a certain order superposed on the violet line (405 nm) of the next higher order. How many lines per cm are there in the grating if the angle of diffraction is 30°?

## Short answer

Equate n×540=(n+1)×405 gives n=3. Grating spacing d=3×540e−9/sin30°=3.24e−6 m. Lines per cm N=1/d=3086 lines/cm.

## Answer

A line of wavelength λ₁ in order *n* is superposed on a line of λ₂ in order (n+1) when the same angle θ satisfies the grating equation for both: d sin θ = nλ₁ = (n+1)λ₂.

**Given:** λ₁ = 540 nm, λ₂ = 405 nm, θ = 30°.

1. Equate: n × 540 = (n + 1) × 405
2. 540n − 405n = 405, so 135n = 405, **n = 3**. The 540 nm line is third order; the 405 nm line is fourth order.
3. Grating element: d sin θ = nλ₁

```text
d = nλ₁ / sin θ = 3 × 540×10⁻⁹ / sin 30°
  = 1.62×10⁻⁶ / 0.5
  = 3.24×10⁻⁶ m = 3.24×10⁻⁴ cm
```

4. Number of lines per cm:

```text
N = 1/d = 1 / (3.24×10⁻⁴ cm) = 3086 lines/cm
```

**Answer: about 3.09 × 10³ lines per cm (≈ 3086 lines/cm).**

## Check yourself

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

1. For λ1 in order n to coincide with λ2 in order n+1 at the same angle, we need
   A. nλ1 = (n+1)λ2
   B. nλ2 = (n+1)λ1
   C. λ1 = λ2
   D. n = n+1
   Answer: A. Both satisfy d sin θ, so nλ1 = (n+1)λ2. (Key point: Condition for superposition)

2. For 540 nm and 405 nm, the order n of the 540 nm line is
   A. 2
   B. 4
   C. 3
   D. 1
   Answer: C. 540n = 405(n+1) gives 135n = 405, so n = 3. (Key point: Order)

3. The grating element d is
   A. 1.62×10⁻⁶ m
   B. 3.24×10⁻⁶ m
   C. 6.48×10⁻⁶ m
   D. 9.0×10⁻⁷ m
   Answer: B. d = nλ1/sin 30° = 3 × 540×10⁻⁹/0.5 = 3.24×10⁻⁶ m. (Key point: Grating element)

4. The number of lines per cm is about
   A. 309
   B. 30860
   C. 6170
   D. 3086
   Answer: D. N = 1/d = 1/(3.24×10⁻⁴ cm) ≈ 3086 lines/cm. (Key point: Lines per cm)

## Related in Engineering Physics

- [Total lines on a grating](https://www.bctengineer.com/question.md?key=semester_1%2Fengineering_physics%3Agrating-total-lines-second-order) — Engineering Physics, chapter 4
- [Dispersive and resolving power of grating](https://www.bctengineer.com/question.md?key=semester_1%2Fengineering_physics%3Agrating-dispersive-resolving-power) — Engineering Physics, chapter 4, asked 5×
- [Minimum lines to resolve sodium lines](https://www.bctengineer.com/question.md?key=semester_1%2Fengineering_physics%3Agrating-minimum-lines-resolve-sodium) — Engineering Physics, chapter 4
- [Sound energy flow per square cm](https://www.bctengineer.com/question.md?key=semester_1%2Fengineering_physics%3Asound-energy-flow-frequency-amplitude) — Engineering Physics, chapter 3
- [Image by two lenses 30 cm and 40 cm](https://www.bctengineer.com/question.md?key=semester_1%2Fengineering_physics%3Atwo-lenses-thirty-forty-image) — Engineering Physics, chapter 5, asked 3×

## Related in other subjects

- [Techniques for higher order active filters](https://www.bctengineer.com/question.md?key=semester_5%2Ffilter_design%3Ahigher-order-active-techniques) — Filter Design, chapter 8
- [Why decompose higher-order filters; ladder realization](https://www.bctengineer.com/question.md?key=semester_6%2Fdigital_control_system%3Adecomposition-ladder-scheme) — Digital Control System, chapter 4
- [Normal incidence, reflection and transmission](https://www.bctengineer.com/question.md?key=semester_3%2Felectromagnetism%3Anormal-incidence-reflection-transmission) — Electromagnetism, chapter 4
- [Normal incidence on perfect conductor](https://www.bctengineer.com/question.md?key=semester_3%2Felectromagnetism%3Anormal-incidence-conducting-plane-100mhz) — Electromagnetism, chapter 4