The Wavelength Formula gives the relationship between the speed of a wave, its frequency, and the distance between two successive crests or troughs, expressed as λ = v/f. This fundamental formula is covered in NCERT Class 11 Physics (Chapter 15 — Waves) and Class 12 Physics (Chapter 10 — Wave Optics). It is equally critical for JEE Main, JEE Advanced, and NEET, where wave-related problems appear every year. This article covers the complete wavelength formula, its derivation, a full physics formula sheet, three progressive solved examples, CBSE exam tips, common mistakes, and JEE/NEET applications.
Key Wavelength Formulas at a Glance
Quick reference for the most important wavelength and wave formulas.
- Basic wavelength formula: \( \lambda = \dfrac{v}{f} \)
- Wave speed relation: \( v = f\lambda \)
- Angular wavenumber: \( k = \dfrac{2\pi}{\lambda} \)
- De Broglie wavelength: \( \lambda = \dfrac{h}{mv} \)
- Photon wavelength (energy): \( E = \dfrac{hc}{\lambda} \)
- Time period relation: \( \lambda = vT \)
- Frequency from wavelength: \( f = \dfrac{v}{\lambda} \)
What is the Wavelength Formula?
The Wavelength Formula defines wavelength as the spatial distance between two consecutive points that are in the same phase of a wave cycle. In simpler terms, it is the distance from one crest to the next crest, or from one trough to the next trough. Wavelength is denoted by the Greek letter lambda (λ) and is measured in metres (m) in the SI system.
According to NCERT Class 11 Physics, Chapter 15 (Waves), every periodic wave has three defining characteristics: speed (v), frequency (f), and wavelength (λ). These three quantities are connected by the fundamental wave equation. The wavelength formula is applicable to all types of waves — mechanical waves (sound, water waves), electromagnetic waves (light, radio waves), and matter waves (de Broglie waves).
The concept also appears in NCERT Class 12 Physics, Chapter 10 (Wave Optics), where the wavelength of visible light determines colour, and in Chapter 11 (Dual Nature of Radiation and Matter), where the de Broglie wavelength links particle momentum to wave behaviour. Understanding the wavelength formula is therefore essential across multiple chapters and examination levels.
Wavelength Formula — Expression and Variables
The primary wavelength formula relates wave speed, frequency, and wavelength:
\[ \lambda = \frac{v}{f} \]
Equivalently, the same relationship is written as the wave equation:
\[ v = f\lambda \]
A second important form uses the time period \( T \) of the wave:
\[ \lambda = vT \]
Since \( T = \dfrac{1}{f} \), both forms are mathematically identical.
| Symbol | Quantity | SI Unit |
|---|---|---|
| \( \lambda \) | Wavelength | Metre (m) |
| \( v \) | Wave speed (phase velocity) | Metre per second (m/s) |
| \( f \) | Frequency | Hertz (Hz) |
| \( T \) | Time period | Second (s) |
| \( k \) | Angular wavenumber | Radian per metre (rad/m) |
| \( h \) | Planck’s constant | Joule-second (J·s) |
| \( m \) | Mass of particle (de Broglie) | Kilogram (kg) |
| \( c \) | Speed of light in vacuum | 3 × 10&sup8; m/s |
Derivation of the Wavelength Formula
Consider a wave travelling with speed \( v \) and frequency \( f \). In one complete oscillation (one time period \( T \)), the wave advances by exactly one wavelength \( \lambda \).
Step 1: By definition, speed = distance / time. In one period, distance = \( \lambda \) and time = \( T \).
\[ v = \frac{\lambda}{T} \]
Step 2: Since frequency is the reciprocal of time period, \( f = \dfrac{1}{T} \), substitute:
\[ v = \lambda \times f \]
Step 3: Rearrange to isolate wavelength:
\[ \lambda = \frac{v}{f} \]
This derivation is direct and elegant. It applies to any wave — sound, light, or water — as long as the wave travels at a well-defined speed.
Complete Physics Formula Sheet — Wavelength and Wave Equations
| Formula Name | Expression | Variables | SI Units | NCERT Chapter |
|---|---|---|---|---|
| Basic Wavelength Formula | \( \lambda = v/f \) | λ = wavelength, v = speed, f = frequency | m | Class 11, Ch 15 |
| Wave Equation | \( v = f\lambda \) | v = wave speed, f = frequency, λ = wavelength | m/s | Class 11, Ch 15 |
| Wavelength via Time Period | \( \lambda = vT \) | λ = wavelength, v = speed, T = time period | m | Class 11, Ch 15 |
| Angular Wavenumber | \( k = 2\pi/\lambda \) | k = wavenumber, λ = wavelength | rad/m | Class 11, Ch 15 |
| De Broglie Wavelength | \( \lambda = h/(mv) \) | h = Planck’s constant, m = mass, v = velocity | m | Class 12, Ch 11 |
| De Broglie via Momentum | \( \lambda = h/p \) | h = Planck’s constant, p = momentum | m | Class 12, Ch 11 |
| Photon Energy & Wavelength | \( E = hc/\lambda \) | E = energy, h = Planck’s constant, c = speed of light | J | Class 12, Ch 11 |
| Speed of Sound in Air | \( v = \sqrt{\gamma P/\rho} \) | γ = adiabatic index, P = pressure, ρ = density | m/s | Class 11, Ch 15 |
| Speed of Light in Medium | \( v = c/n \) | c = speed of light in vacuum, n = refractive index | m/s | Class 12, Ch 10 |
| Wavelength in Medium | \( \lambda_m = \lambda_0/n \) | λ⊂m; = wavelength in medium, λ⊂0; = wavelength in vacuum, n = refractive index | m | Class 12, Ch 10 |
| Young’s Double Slit Fringe Width | \( \beta = \lambda D/d \) | β = fringe width, D = screen distance, d = slit separation | m | Class 12, Ch 10 |
Wavelength Formula — Solved Examples
Example 1 (Class 9–10 Level): Finding Wavelength of a Sound Wave
Problem: A sound wave travels through air at a speed of 340 m/s. Its frequency is 680 Hz. Calculate the wavelength of the sound wave.
Given:
- Speed of sound, \( v = 340 \) m/s
- Frequency, \( f = 680 \) Hz
Step 1: Write the wavelength formula: \( \lambda = \dfrac{v}{f} \)
Step 2: Substitute the known values:
\[ \lambda = \frac{340}{680} \]
Step 3: Simplify the expression:
\[ \lambda = 0.5 \text{ m} \]
Answer
The wavelength of the sound wave is 0.5 m (50 cm).
Example 2 (Class 11–12 Level): Wavelength of Light in a Medium
Problem: Yellow light has a wavelength of 589 nm in vacuum. It enters a glass medium with a refractive index of 1.5. Find (a) the speed of light in glass and (b) the wavelength of yellow light in glass.
Given:
- Wavelength in vacuum, \( \lambda_0 = 589 \) nm \( = 589 \times 10^{-9} \) m
- Refractive index of glass, \( n = 1.5 \)
- Speed of light in vacuum, \( c = 3 \times 10^8 \) m/s
Step 1: Find the speed of light in glass using \( v = c/n \):
\[ v = \frac{3 \times 10^8}{1.5} = 2 \times 10^8 \text{ m/s} \]
Step 2: Find the frequency of yellow light (frequency does not change in a medium):
\[ f = \frac{c}{\lambda_0} = \frac{3 \times 10^8}{589 \times 10^{-9}} \approx 5.09 \times 10^{14} \text{ Hz} \]
Step 3: Find the wavelength in glass using \( \lambda_m = \lambda_0 / n \):
\[ \lambda_m = \frac{589}{1.5} \approx 392.7 \text{ nm} \]
Verification: Using \( \lambda_m = v/f \): \( \lambda_m = (2 \times 10^8) / (5.09 \times 10^{14}) \approx 393 \) nm. Both methods agree.
Answer
(a) Speed of light in glass = 2 × 10&sup8; m/s. (b) Wavelength in glass ≈ 392.7 nm.
Example 3 (JEE/NEET Level): De Broglie Wavelength of an Electron
Problem: An electron is accelerated through a potential difference of 100 V. Calculate the de Broglie wavelength associated with the electron. (Given: mass of electron \( m_e = 9.1 \times 10^{-31} \) kg, charge \( e = 1.6 \times 10^{-19} \) C, Planck’s constant \( h = 6.626 \times 10^{-34} \) J·s)
Given:
- Potential difference, \( V = 100 \) V
- \( m_e = 9.1 \times 10^{-31} \) kg
- \( e = 1.6 \times 10^{-19} \) C
- \( h = 6.626 \times 10^{-34} \) J·s
Step 1: Find the kinetic energy gained by the electron:
\[ KE = eV = 1.6 \times 10^{-19} \times 100 = 1.6 \times 10^{-17} \text{ J} \]
Step 2: Use \( KE = p^2 / (2m) \) to find momentum \( p \):
\[ p = \sqrt{2m_e \cdot KE} = \sqrt{2 \times 9.1 \times 10^{-31} \times 1.6 \times 10^{-17}} \]
\[ p = \sqrt{2.912 \times 10^{-47}} \approx 5.396 \times 10^{-24} \text{ kg m/s} \]
Step 3: Apply the de Broglie wavelength formula \( \lambda = h/p \):
\[ \lambda = \frac{6.626 \times 10^{-34}}{5.396 \times 10^{-24}} \approx 1.228 \times 10^{-10} \text{ m} \]
Step 4: Convert to Ångströms for convenience: \( \lambda \approx 1.23 \) Å (1 Å = 10&sup-10; m).
A useful shortcut for this type of problem: \( \lambda = \dfrac{12.27}{\sqrt{V}} \) Å, where V is in volts. Checking: \( 12.27 / \sqrt{100} = 12.27/10 = 1.227 \) Å. This matches our detailed calculation perfectly.
Answer
The de Broglie wavelength of the electron is approximately 1.23 Å (1.23 × 10&sup-10; m).
CBSE Exam Tips 2025-26
- Memorise the three forms: Learn \( v = f\lambda \), \( \lambda = v/f \), and \( \lambda = vT \) as a trio. CBSE often asks you to rearrange and find any one of the three variables.
- Unit consistency is crucial: Always convert frequency to Hz and speed to m/s before substituting. A common error is using kHz or MHz without converting, which gives wrong answers.
- Electromagnetic spectrum values: CBSE Class 12 board exams frequently ask about the wavelength ranges of radio waves, microwaves, infrared, visible light, UV, X-rays, and gamma rays. Memorise these ranges from NCERT Table 8.1.
- De Broglie shortcut: For electrons accelerated through potential V, use \( \lambda = 12.27/\sqrt{V} \) Å. We recommend memorising this shortcut — it saves valuable time in CBSE and competitive exams.
- Frequency is invariant: Remember that when light travels from one medium to another, its frequency remains constant. Only speed and wavelength change. This is a very common CBSE MCQ trap.
- Young’s Double Slit: The fringe width formula \( \beta = \lambda D/d \) directly involves wavelength. Practice rearranging it to find \( \lambda \) from given fringe width data — this is a standard 3-mark question in 2025-26 board exams.
Common Mistakes to Avoid with the Wavelength Formula
- Confusing frequency and angular frequency: The wavelength formula uses ordinary frequency \( f \) (in Hz), not angular frequency \( \omega \) (in rad/s). The relation \( \omega = 2\pi f \) must be used to convert. Students who substitute \( \omega \) directly into \( \lambda = v/f \) get an answer that is \( 2\pi \) times too small.
- Using the wrong wave speed: Sound travels at approximately 340 m/s in air at room temperature. Light travels at \( 3 \times 10^8 \) m/s in vacuum. Always identify the medium and wave type before choosing the correct speed value.
- Ignoring refractive index changes: When light enters a denser medium, its speed and wavelength both decrease by a factor of \( n \). Many students forget to apply \( \lambda_m = \lambda_0/n \) and incorrectly use the vacuum wavelength throughout.
- Incorrect unit conversion for nanometres: Visible light wavelengths are given in nanometres (nm). Remember that 1 nm = \( 10^{-9} \) m. Forgetting this conversion leads to enormous errors in energy calculations using \( E = hc/\lambda \).
- Mixing up wavelength and amplitude: Wavelength is a spatial property (distance between crests). Amplitude is the maximum displacement from equilibrium. These are completely different quantities. CBSE MCQs sometimes test this distinction directly.
JEE/NEET Application of the Wavelength Formula
In our experience, JEE aspirants encounter the wavelength formula in at least three to four questions per paper across different contexts. Understanding these application patterns gives you a significant advantage.
Pattern 1: De Broglie Wavelength in Modern Physics (JEE Main & NEET)
Both JEE and NEET frequently ask about the de Broglie wavelength \( \lambda = h/p = h/mv \) for electrons, protons, and alpha particles. A classic question type compares the de Broglie wavelengths of particles with the same kinetic energy. Since \( \lambda = h/\sqrt{2mKE} \), the particle with greater mass has a shorter wavelength. For equal kinetic energy, the ratio of wavelengths is \( \lambda_1/\lambda_2 = \sqrt{m_2/m_1} \). This ratio-based approach appears very commonly in JEE Main MCQs.
Pattern 2: Young’s Double Slit and Wavelength (JEE Advanced)
JEE Advanced problems combine the wavelength formula with interference optics. Given the fringe width \( \beta \), slit separation \( d \), and screen distance \( D \), you must find \( \lambda \) using \( \lambda = \beta d / D \). More complex problems involve two wavelengths overlapping at certain fringes, requiring you to find the condition \( n_1 \lambda_1 = n_2 \lambda_2 \). Mastering the wavelength formula is the first step to solving these multi-concept problems.
Pattern 3: Photon Energy and Threshold Wavelength (NEET)
NEET questions on the photoelectric effect heavily use the relation \( E = hc/\lambda \). You are typically asked to find the threshold wavelength \( \lambda_0 = hc/\phi \), where \( \phi \) is the work function. Alternatively, you may need to find the maximum kinetic energy of emitted electrons: \( KE_{max} = hc/\lambda – hc/\lambda_0 \). In our experience, NEET aspirants who are comfortable rearranging the photon wavelength formula solve these questions in under 60 seconds.
Pattern 4: Stationary Waves and Resonance (JEE Main)
In standing wave problems, the wavelength formula connects to resonance conditions. For a string of length \( L \) fixed at both ends, the allowed wavelengths are \( \lambda_n = 2L/n \) for \( n = 1, 2, 3, \ldots \) The corresponding frequencies are \( f_n = nv/(2L) \). JEE Main asks you to identify harmonic numbers and calculate frequencies from given wavelengths. Knowing \( v = f\lambda \) is the anchor for all such calculations.
FAQs on Wavelength Formula
Explore More Physics Formulas
We hope this comprehensive guide to the Wavelength Formula has helped you build a clear understanding of wave behaviour across all levels. To strengthen your preparation further, explore these related formula articles on ncertbooks.net:
- Learn how energy is transmitted by waves with the Wave Power Formula — essential for understanding wave intensity in JEE problems.
- Understand energy transfer in thermal processes with the Heat Transfer Formula, which complements wave energy concepts in thermodynamics.
- Explore the quantised energy of light particles with the Photon Energy Formula (\( E = hf = hc/\lambda \)) — directly linked to the wavelength formula in modern physics.
- Browse our complete Physics Formulas hub for the full list of Class 11 and Class 12 formula articles, organised by chapter.
For official NCERT textbook content and syllabus updates, refer to the NCERT official website (ncert.nic.in).