---
title: "Dispersive and resolving power of grating"
subject: "Engineering Physics"
chapter: 4
key: "semester_1/engineering_physics:grating-dispersive-resolving-power"
source: https://www.bctengineer.com/q/semester_1/engineering_physics/grating-dispersive-resolving-power
site: BCT Engineering
---

# Dispersive and resolving power of grating

Engineering Physics · Chapter 4: Physical Optics

- Asked in: 2074 Chaitra, 5 marks; 2073 Shrawan, 5 marks; 2071 Chaitra, 5 marks; 2070 Asar, 5 marks; 2068 Chaitra, 5 marks (asked 5 times)
- Video: [Resolving Power and Dispersive Power of a Grating](https://www.youtube.com/watch?v=OnYiflzz2n4) — Physics by Grace
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_1/engineering_physics/grating-dispersive-resolving-power
- Whole chapter: https://www.bctengineer.com/md/semester_1/engineering_physics/chapter-4

## Question

What is diffraction of light? Explain the dispersive power and resolving power of a diffraction grating. Derive their expressions and relate them (show that the resolving power equals the total number of rulings times the order).

## Short answer

Diffraction is bending of light around small apertures. Dispersive power dθ/dλ = n/(d cosθ). Resolving power R = λ/Δλ = nN (N total rulings), which equals (dispersive power)×(W cosθ).

## Answer

**Diffraction of light** is the bending of light around the edges of an obstacle or aperture whose size is comparable to the wavelength, so that light enters the geometrical shadow and bright and dark fringes form.

A **plane transmission grating** has many equal parallel slits of width *a* separated by opaque spaces *b*; *d* = a + b is the grating element. Maxima of order *n* occur at

```text
d sin θ = n λ        (n = 0, 1, 2 ...)         (1)
```

#### Dispersive power
It is the rate of change of the diffraction angle with wavelength, dθ/dλ.
1. Differentiate (1) with respect to λ: d cos θ · dθ = n · dλ
2. Therefore **dθ/dλ = n / (d cos θ)**

So the dispersion is larger for a higher order, a smaller grating element (more lines per cm) and a larger θ.

#### Resolving power
It is R = λ/dλ, where dλ is the smallest wavelength difference between two lines (λ and λ + dλ) that are just resolved by the Rayleigh criterion (the principal maximum of one falls on the first minimum of the other).
1. Let the grating have N lines. The *n*th maximum of λ + dλ is at: d sin θ = n(λ + dλ)
2. The first minimum next to the *n*th maximum of λ occurs when the path difference across the whole grating is (nN + 1)λ: N d sin θ = (nN + 1)λ
3. Multiply step 1 by N: N d sin θ = nN(λ + dλ)
4. Equate: nN(λ + dλ) = nNλ + λ, so nN dλ = λ
5. Therefore **R = λ/dλ = nN**

#### Relation between them
Using N d = total width W of the ruled part of the grating:

```text
R = nN = (n / d cos θ) × (N d cos θ) = (dθ/dλ) × W cos θ
```

So resolving power = (dispersive power) × (width of the diffracted beam, W cos θ). Also R = nN: the resolving power equals the total number of rulings times the order. A wide grating with many lines, used in a higher order, resolves closer wavelengths. For example, in the first order (n = 1) a grating with N = 15000 lines has R = λ/dλ = 15000.

## Check yourself

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

1. Principal maxima of a grating satisfy
   A. d sin θ = nλ
   B. d cos θ = nλ
   C. a sin θ = nλ
   D. d sin θ = λ/n
   Answer: A. With grating element d = a + b, maxima occur at d sin θ = nλ. (Key point: Grating equation)

2. Differentiating d sin θ = nλ gives dispersive power dθ/dλ =
   A. n d cos θ
   B. d/(n cos θ)
   C. n/(d cos θ)
   D. n sin θ/d
   Answer: C. d cos θ dθ = n dλ, so dθ/dλ = n/(d cos θ). (Key point: Dispersive power)

3. The resolving power of a grating with N lines in order n is
   A. n/N
   B. N/n
   C. n + N
   D. nN
   Answer: D. Using the Rayleigh criterion, nN dλ = λ, so R = λ/dλ = nN. (Key point: Resolving power)

4. Two lines are just resolved when
   A. their maxima coincide exactly
   B. the principal maximum of one falls on the first minimum of the other
   C. they lie in different orders
   D. their intensities are equal
   Answer: B. That is the Rayleigh criterion used to find the smallest resolvable dλ. (Key point: Rayleigh criterion)

## 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
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- [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×
- [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×

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