The Diffraction Grating Formula is expressed as \ ( d\sin\theta = n\lambda \), and it describes how a diffraction grating separates light into its component wavelengths. This formula is a cornerstone of wave optics, covered in Class 12 Physics (NCERT Chapter 10 — Wave Optics). It is also a high-weightage topic in JEE Main, JEE Advanced, and NEET. This article covers the complete derivation, a formula sheet, three progressive solved examples, CBSE exam tips, common mistakes, and FAQs — everything you need in one place.
Key Diffraction Grating Formulas at a Glance
Quick reference for the most important diffraction grating formulas.
- Grating equation: \( d\sin\theta = n\lambda \)
- Grating element: \( d = \frac{1}{N} \) where N = lines per metre
- Maximum order: \( n_{\max} = \frac{d}{\lambda} \)
- Angular dispersion: \( \frac{d\theta}{d\lambda} = \frac{n}{d\cos\theta} \)
- Resolving power: \( R = nN \)
- Rayleigh criterion for gratings: \( \Delta\lambda = \frac{\lambda}{nN} \)
- Path difference condition: \( \Delta = d\sin\theta \)
What is the Diffraction Grating Formula?
The Diffraction Grating Formula gives the condition for constructive interference when light passes through (or reflects off) a diffraction grating. A diffraction grating is an optical device with a large number of equally spaced parallel slits or rulings. When monochromatic or polychromatic light falls on a grating, each slit acts as a secondary source of waves. These waves interfere constructively at specific angles, producing bright fringes called principal maxima.
The formula is studied in NCERT Class 12 Physics, Chapter 10 (Wave Optics). It builds directly on the concepts of Huygens’ principle and Young’s double-slit experiment. The key insight is simple: constructive interference occurs whenever the path difference between adjacent slits equals an integer multiple of the wavelength.
In competitive exams such as JEE Main and NEET, this formula is used to calculate the wavelength of light, identify spectral orders, and analyse resolving power. Understanding the Diffraction Grating Formula thoroughly is essential for scoring full marks in the wave optics section.
Diffraction Grating Formula — Expression and Variables
The standard diffraction grating equation for the n-th order principal maximum is:
\[ d\sin\theta = n\lambda \]
Here, \( d \) is the grating element (slit spacing), \( \theta \) is the diffraction angle, \( n \) is the order of diffraction, and \( \lambda \) is the wavelength of light.
| Symbol | Quantity | SI Unit |
|---|---|---|
| \( d \) | Grating element (distance between adjacent slits) | metre (m) |
| \( \theta \) | Angle of diffraction for n-th order maximum | degree or radian |
| \( n \) | Order of diffraction (integer: 0, ±1, ±2, …) | dimensionless |
| \( \lambda \) | Wavelength of incident light | metre (m) or nm |
| \( N \) | Number of lines (rulings) per unit length | lines per metre (m⁻¹) |
| \( R \) | Resolving power of the grating | dimensionless |
Derivation of the Diffraction Grating Formula
Consider a plane diffraction grating with N slits, each separated by a distance \( d \) (the grating element). A monochromatic light beam of wavelength \( \lambda \) is incident normally on the grating.
Step 1: Each slit diffracts light in all forward directions. Consider two adjacent slits, A and B, separated by distance \( d \).
Step 2: For light diffracted at angle \( \theta \) to the normal, the path difference between rays from A and B is \( \Delta = d\sin\theta \).
Step 3: Constructive interference (bright fringe) occurs when the path difference equals an integer multiple of the wavelength:
\[ \Delta = n\lambda \quad \Rightarrow \quad d\sin\theta = n\lambda \]
Step 4: Here, \( n = 0, \pm1, \pm2, \ldots \) represents the zeroth, first, second, … orders of diffraction. The maximum possible order is limited by \( \sin\theta \leq 1 \), giving \( n_{\max} = \lfloor d/\lambda \rfloor \).
This derivation applies to a transmission grating. For a reflection grating, the same equation holds because the geometry of path differences is identical.
Complete Wave Optics Formula Sheet
| Formula Name | Expression | Variables | SI Units | NCERT Chapter |
|---|---|---|---|---|
| Diffraction Grating (principal maxima) | \( d\sin\theta = n\lambda \) | d = grating element, n = order, λ = wavelength | m, dimensionless | Class 12, Ch 10 |
| Grating Element | \( d = \frac{1}{N} \) | N = number of lines per metre | m | Class 12, Ch 10 |
| Maximum Diffraction Order | \( n_{\max} = \left\lfloor \frac{d}{\lambda} \right\rfloor \) | d = grating element, λ = wavelength | dimensionless | Class 12, Ch 10 |
| Angular Dispersion | \( \frac{d\theta}{d\lambda} = \frac{n}{d\cos\theta} \) | n = order, d = grating element, θ = diffraction angle | rad/m | Class 12, Ch 10 |
| Resolving Power | \( R = nN \) | n = order, N = total number of slits | dimensionless | Class 12, Ch 10 |
| Rayleigh Resolution Limit | \( \Delta\lambda = \frac{\lambda}{nN} \) | λ = wavelength, n = order, N = number of slits | m | Class 12, Ch 10 |
| Single-Slit Diffraction Minima | \( a\sin\theta = m\lambda \) | a = slit width, m = 1, 2, 3…, λ = wavelength | m | Class 12, Ch 10 |
| Young’s Double Slit (fringe width) | \( \beta = \frac{\lambda D}{d} \) | λ = wavelength, D = screen distance, d = slit separation | m | Class 12, Ch 10 |
| Brewster’s Law | \( \tan\theta_B = \mu \) | θ_B = Brewster angle, μ = refractive index | dimensionless | Class 12, Ch 10 |
| Malus’ Law | \( I = I_0\cos^2\theta \) | I = transmitted intensity, I₀ = incident intensity, θ = angle | W/m² | Class 12, Ch 10 |
Diffraction Grating Formula — Solved Examples
Example 1 (Class 9-10 Level — Direct Application)
Problem: A diffraction grating has 500 lines per mm. Monochromatic light of wavelength 600 nm falls normally on it. Find the angle of diffraction for the first-order maximum.
Given:
- Number of lines per mm: 500, so lines per metre N = 500 × 10³ = 5 × 10⁵ m⁻¹
- Grating element: \( d = \frac{1}{N} = \frac{1}{5 \times 10^5} = 2 \times 10^{-6} \) m
- Wavelength: \( \lambda = 600 \) nm \( = 600 \times 10^{-9} \) m
- Order: \( n = 1 \)
Step 1: Write the Diffraction Grating Formula: \( d\sin\theta = n\lambda \)
Step 2: Substitute values:
\[ \sin\theta = \frac{n\lambda}{d} = \frac{1 \times 600 \times 10^{-9}}{2 \times 10^{-6}} = \frac{6 \times 10^{-7}}{2 \times 10^{-6}} = 0.30 \]
Step 3: Find the angle: \( \theta = \sin^{-1}(0.30) \approx 17.46^\circ \)
Answer
The angle of diffraction for the first-order maximum is approximately 17.46°.
Example 2 (Class 11-12 Level — Multi-Step)
Problem: A grating with 4000 lines per cm is illuminated normally with white light. Determine (a) the highest order visible for wavelength 700 nm (red), and (b) whether the second-order red and third-order violet (λ = 400 nm) spectra overlap.
Given:
- Lines per cm: 4000, so N = 4000 × 100 = 4 × 10⁵ m⁻¹
- Grating element: \( d = \frac{1}{4 \times 10^5} = 2.5 \times 10^{-6} \) m
- \( \lambda_{\text{red}} = 700 \) nm \( = 7 \times 10^{-7} \) m
- \( \lambda_{\text{violet}} = 400 \) nm \( = 4 \times 10^{-7} \) m
Step 1: Find the maximum order for red light. Since \( \sin\theta \leq 1 \):
\[ n_{\max} = \frac{d}{\lambda} = \frac{2.5 \times 10^{-6}}{7 \times 10^{-7}} \approx 3.57 \]
So \( n_{\max} = 3 \) (taking the integer part).
Step 2: Check if 2nd-order red overlaps with 3rd-order violet. Two wavelengths overlap in different orders when \( n_1\lambda_1 = n_2\lambda_2 \).
\[ n_1\lambda_1 = 2 \times 700 = 1400 \text{ nm} \]
\[ n_2\lambda_2 = 3 \times 400 = 1200 \text{ nm} \]
Step 3: Since 1400 nm ≠ 1200 nm, the two spectra do not overlap at the same angle. They appear at different angles.
Answer
(a) Highest visible order for red light = 3rd order. (b) The 2nd-order red and 3rd-order violet spectra do not overlap.
Example 3 (JEE/NEET Level — Concept Application)
Problem: A diffraction grating has a total of 6000 slits, each of width 0.8 μm, and a grating element d = 2.0 μm. Light of wavelength 500 nm is incident normally. (a) Find the angles for the 1st and 2nd order principal maxima. (b) Determine the resolving power in the 2nd order. (c) Find the minimum wavelength difference that can be resolved in the 2nd order.
Given:
- Total slits: N = 6000
- Grating element: \( d = 2.0 \times 10^{-6} \) m
- Wavelength: \( \lambda = 500 \) nm \( = 5 \times 10^{-7} \) m
Step 1: Find angles using \( d\sin\theta = n\lambda \).
For n = 1:
\[ \sin\theta_1 = \frac{\lambda}{d} = \frac{5 \times 10^{-7}}{2 \times 10^{-6}} = 0.25 \quad \Rightarrow \quad \theta_1 = 14.48^\circ \]
For n = 2:
\[ \sin\theta_2 = \frac{2\lambda}{d} = \frac{2 \times 5 \times 10^{-7}}{2 \times 10^{-6}} = 0.50 \quad \Rightarrow \quad \theta_2 = 30.00^\circ \]
Step 2: Resolving power in 2nd order: \( R = nN = 2 \times 6000 = 12000 \).
Step 3: Minimum resolvable wavelength difference:
\[ \Delta\lambda = \frac{\lambda}{R} = \frac{500}{12000} \approx 0.0417 \text{ nm} \]
Answer
(a) θ₁ ≈ 14.48°, θ₂ = 30.00°. (b) Resolving power = 12,000. (c) Minimum resolvable wavelength difference ≈ 0.042 nm.
CBSE Exam Tips 2025-26
- Always convert units first. Wavelength is often given in nm or Å. Convert to metres before substituting into \( d\sin\theta = n\lambda \). A unit error alone can cost you full marks.
- State the formula clearly. In CBSE 2025-26 board exams, writing the formula before substitution earns a dedicated step mark. Never skip it.
- Remember the grating element relation. If a question gives “lines per mm” or “lines per cm,” first compute \( d = 1/N \) in metres. We recommend practising this conversion until it is automatic.
- Know the maximum order condition. Questions frequently ask for the highest visible order. Use \( n_{\max} = \lfloor d/\lambda \rfloor \), because \( \sin\theta \) cannot exceed 1.
- Distinguish grating from single-slit diffraction. The grating formula \( d\sin\theta = n\lambda \) gives maxima. The single-slit formula \( a\sin\theta = m\lambda \) gives minima. Confusing the two is a common board-exam error.
- Practise resolving power numericals. The formula \( R = nN \) appears in 3-mark questions in recent CBSE sample papers. It is easy marks if you know it.
Common Mistakes to Avoid
- Mistake 1 — Wrong unit for N. Students often use “lines per mm” directly as N without converting to “lines per metre.” Always convert: 600 lines/mm = 6 × 10⁵ lines/m, giving d = 1/(6 × 10⁵) m.
- Mistake 2 — Confusing order with number of slits. The variable \( n \) in the grating formula is the diffraction order (0, 1, 2, …), not the number of slits N. These are two completely different quantities.
- Mistake 3 — Ignoring the maximum order constraint. Students sometimes calculate \( \sin\theta > 1 \) and report a non-physical answer. Always check: if \( \sin\theta > 1 \), that order does not exist.
- Mistake 4 — Applying the grating formula to minima. The formula \( d\sin\theta = n\lambda \) gives principal maxima only. Minima of the grating pattern have a different (more complex) condition.
- Mistake 5 — Forgetting the zeroth order. The zeroth order (n = 0) corresponds to \( \theta = 0 \), i.e., straight-through light. It carries no spectral information. Do not confuse it with a missing order.
JEE/NEET Application of the Diffraction Grating Formula
In our experience, JEE aspirants encounter the Diffraction Grating Formula in at least one question per year in JEE Main, and occasionally in JEE Advanced as part of a multi-concept optics problem. NEET typically tests the conceptual understanding of grating spectra and wavelength determination.
Application Pattern 1 — Wavelength Determination
JEE Main frequently gives the grating element, the diffraction angle, and the order, then asks for the wavelength. Rearrange the formula directly:
\[ \lambda = \frac{d\sin\theta}{n} \]
This is a one-step calculation. The difficulty usually lies in correctly computing \( d \) from the given “lines per mm” data.
Application Pattern 2 — Overlapping Spectra
JEE Advanced problems ask whether two spectral orders overlap. Two wavelengths \( \lambda_1 \) and \( \lambda_2 \) in orders \( n_1 \) and \( n_2 \) overlap when:
\[ n_1\lambda_1 = n_2\lambda_2 \]
For example, the 2nd order of 600 nm overlaps with the 3rd order of 400 nm because \( 2 \times 600 = 3 \times 400 = 1200 \) nm. This is a classic JEE-style question.
Application Pattern 3 — Resolving Power
Both JEE and NEET test the resolving power formula \( R = nN \). A typical question asks: “How many slits must a grating have to resolve the sodium doublet (589.0 nm and 589.6 nm) in the first order?” Solve using:
\[ N = \frac{R}{n} = \frac{\lambda}{n\,\Delta\lambda} = \frac{589.0}{1 \times 0.6} \approx 982 \text{ slits} \]
Our experts suggest memorising all three application patterns. Together they cover nearly every grating question that appears in competitive exams.
FAQs on Diffraction Grating Formula
For a deeper understanding of related wave optics concepts, explore our detailed guide on the Refractive Index Formula, which explains how light behaves at optical interfaces. You can also review the Uniform Circular Motion Formula for rotational wave analogies, and visit our complete Physics Formulas hub for the full Class 12 formula library. For the official NCERT syllabus, refer to the NCERT official website.