---
title: "Orders for entire visible spectrum"
subject: "Engineering Physics"
chapter: 4
key: "semester_1/engineering_physics:grating-orders-visible-spectrum"
source: https://www.bctengineer.com/q/semester_1/engineering_physics/grating-orders-visible-spectrum
site: BCT Engineering
---

# Orders for entire visible spectrum

Engineering Physics · Chapter 4: Physical Optics

- Asked in: 2073 Chaitra, 5 marks; 2071 Shrawan, 5 marks (asked 2 times)
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_1/engineering_physics/grating-orders-visible-spectrum
- Whole chapter: https://www.bctengineer.com/md/semester_1/engineering_physics/chapter-4

## Question

A grating with 250 grooves/mm is used with an incandescent light source. Assume visible spectrum to range in wavelength from 400 to 700 nm. In how many orders can one see the entire visible spectrum?

## Short answer

Grating spacing d = 1/250 mm = 4000 nm. The entire visible spectrum (400–700 nm) is observed up to the order where the longest wavelength fits: m_max = d/λ_max = 4000/700 ≈ 5. So the entire spectrum appears in 5 orders (m = 1 to 5) on each side.

## Answer

Grating equation: d sin θ = mλ, where m is the order. Order *m* exists only if sin θ ≤ 1, i.e. mλ ≤ d.

**Given:** 250 grooves/mm, so d = 1/250 mm = 4.0×10⁻⁶ m = 4000 nm; λ ranges from 400 nm to 700 nm.

1. Highest order in which the longest wavelength (700 nm) appears: m ≤ d/λ_max = 4000/700 = 5.71. So m = 1 to 5.
2. Highest order for 400 nm: m ≤ 4000/400 = 10 (not the limiting case).
3. The **entire** spectrum (400 to 700 nm) is therefore present only up to m = 5. The 6th order contains only part of the spectrum, from violet up to 4000/6 ≈ 667 nm.
4. Overlap check: the red end of order *m* (mλ = 700m) overlaps the violet end of order m+1 (400(m+1)) if 700m > 400(m+1), i.e. m > 1.33.

| Order | Range of mλ = d sin θ (nm) | Overlap with next? |
|---|---|---|
| 1 | 400 - 700 | No (next starts at 800) |
| 2 | 800 - 1400 | Yes (3rd starts at 1200) |
| 3 | 1200 - 2100 | Yes |

(The angles follow from sin θ = mλ/d; e.g. first-order red is at sin θ = 0.175, θ = 10.1°.)

**Answer: the entire visible spectrum appears in 5 orders (m = 1 to 5) on each side; only the first order is free from overlap with its neighbour.**

## Check yourself

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

1. A grating has 250 grooves/mm. What is the grating spacing d?
   A. 250 nm
   B. 4000 nm
   C. 400 nm
   D. 2500 nm
   Answer: B. d = 1/250 mm = 4×10^-6 m = 4000 nm. (Key point: Grating spacing)

2. Which wavelength limits the highest order in which the whole spectrum is visible?
   A. 400 nm, the shortest
   B. 700 nm, the longest
   C. 550 nm, the middle
   D. Both equally
   Answer: B. Order m needs mλ ≤ d; the longest wavelength (700 nm) fails first, giving m ≤ 4000/700 = 5.71. (Key point: Limiting wavelength)

3. In how many orders does the entire visible spectrum appear?
   A. 10
   B. 6
   C. 5
   D. 1
   Answer: C. m ≤ 5.71, so only m = 1 to 5 contain the full 400-700 nm range. (Key point: Number of complete orders)

4. Which order is free from overlap with the next order?
   A. Only the first order
   B. The first and second orders
   C. Only the fifth order
   D. All orders
   Answer: A. Order 1 ends at 700 nm while order 2 begins at 800 nm; from order 2 onward the red end overlaps the next violet. (Key point: Overlap)

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