What Is the Relationship Between Light Intensity and Wavelength?

Light intensity and wavelength describe different properties. Intensity tells you how much energy arrives at a surface per unit area per unit time; wavelength describes the spatial period of the wave. One does not determine the other unless you also specify what is being held fixed, such as wave amplitude or photon flux.

What “intensity” means

In many physics problems, intensity means irradiance: power incident on a surface divided by its area. Its SI unit is watts per square metre (W/m²). In everyday language, “intensity” can also mean perceived brightness, which depends on the eye’s sensitivity to different wavelengths and is not the same as physical irradiance.

For a sinusoidal electromagnetic plane wave traveling through vacuum, the time-averaged intensity is

$$I = \frac{1}{2}\epsilon_0 c E_0^2,$$

where $E_0$ is the peak electric-field amplitude, $\epsilon_0$ is the vacuum permittivity, and $c$ is the speed of light. This formula contains no wavelength. If the wave’s electric-field amplitude is held fixed, changing its wavelength does not change its average intensity in vacuum.

That does not mean intensity can never vary with wavelength. It means wavelength alone is not enough to determine intensity: the answer depends on the physical constraint and on how the light was produced or filtered.

When intensity is related to wavelength

A photon’s energy is related to its wavelength by

$$E_{\text{photon}} = \frac{hc}{\lambda},$$

where $h$ is Planck’s constant and $\lambda$ is the wavelength. Shorter-wavelength photons carry more energy than longer-wavelength photons.

If $\Phi$ is the photon flux—the number of photons arriving per square metre per second—then the intensity of monochromatic light is

$$I = \Phi E_{\text{photon}} = \Phi\frac{hc}{\lambda}.$$

So, at fixed photon flux, intensity is inversely proportional to wavelength: halving the wavelength doubles the intensity. But at fixed intensity, shorter-wavelength light has a lower photon flux, because each photon carries more energy.

What is held fixed?	Relationship for monochromatic light
Electric-field amplitude in vacuum	Intensity stays the same as wavelength changes
Photon flux	Intensity is proportional to $1/\lambda$
Intensity	Photon flux is proportional to $\lambda$

These comparisons assume the stated quantity really is fixed. A real source may change its amplitude, photon output, or spectrum when its wavelength changes.

Worked example: photon flux at 500 nm

Suppose monochromatic light has an intensity of $100\ \text{W/m}^2$ and a wavelength of $500\ \text{nm}$, or $5.00\times10^{-7}\ \text{m}$. First find the energy per photon:

$$\begin{aligned} E_{\text{photon}} &= \frac{hc}{\lambda} \\ &= \frac{(6.626\times10^{-34}\ \text{J\,s})(2.998\times10^8\ \text{m/s})}{5.00\times10^{-7}\ \text{m}} \\ &\approx 3.97\times10^{-19}\ \text{J}. \end{aligned}$$

Intensity is energy delivered per area per second, so dividing by the energy per photon gives the photon flux:

$$\Phi = \frac{I}{E_{\text{photon}}} \approx \frac{100\ \text{J\,s}^{-1}\text{m}^{-2}}{3.97\times10^{-19}\ \text{J}} \approx 2.52\times10^{20}\ \text{photons\,m}^{-2}\text{s}^{-1}.$$

At the same photon flux, 250 nm light would have twice the intensity of 500 nm light, because each 250 nm photon has twice the energy. This comparison is about photon energy, not a rule that every 250 nm source is more intense than every 500 nm source.

Light with many wavelengths

Most sources are not perfectly monochromatic. Their intensity can be described by a spectral irradiance, such as $S_\lambda$, which gives power per area per wavelength interval. The total irradiance is found by adding across the spectrum:

$$I_{\text{total}} = \int S_\lambda\,d\lambda.$$

The shape of $S_\lambda$ depends on the source and any materials the light passes through. For example, a filter can remove some wavelengths, and a hot object emits a spectrum whose distribution changes with temperature. In these cases, intensity varies across wavelength because of the source or material, not because wavelength universally dictates intensity.

Common mistakes

• Assuming intensity is proportional to wavelength in every situation. The inverse relationship applies when photon flux is fixed; it is not a universal wave law.
• Confusing photon energy with total intensity. Shorter wavelength means more energy per photon, but total intensity also depends on how many photons arrive.
• Treating visible brightness as physical intensity. Human vision responds differently to different wavelengths, so equal irradiance at two wavelengths may not look equally bright.
• Using one wavelength for a broad spectrum. For broadband light, account for the power at each wavelength and integrate to get total intensity.