# Is the intensity of a light wave related to frequency of the wave?

My problem is: How can I resolve these following ideas?

1. Energy of photons in an EM wave is proportional to the frequency of the wave

2. Intensity of an EM wave is proportional to the energy that is incident an area per unit time

3. Intensity is proportional to the number of photons incident on an area per unit time

Based on reviewing previous physics stack exchange questions (such as Relation between frequency and intensity of light):

Intensity = I = Nhν/A if N is the monochromatic photon emission rate (photons per second), ν is the frequency of the photons, and A is the area these photons are hitting.

The above equation makes sense to me.

However, I'm being told that intensity of an EM wave is not related to the frequency of the wave because "light intensity is a wave-like property that depends on the number of photons hitting a certain area per unit time. Conceptualizing it into the units of power, the Joules component refers to the kinetic energy of photons (1/2 * mv^2). Since light travels at the same speed in a given medium, regardless of frequency, each individual photon contributes equally to energy, and thus to intensity".

But if there are two waves with equal numbers of photons incident on an area per unit time, but one wave has greater frequency, then wouldn't the wave with greater frequency have more energy incident on the area per unit time?

Can I resolve my confusion by using (energy = 1/2 * mv^2)?

Thank you for any help in understanding this!

## 2 Answers

It might help to think of throwing rocks instead of photons. Pretend blue light is made of big blue rocks. When you throw them at a target, each rock carries a lot of energy.

Red light is made of smaller red rocks. When you throw them at a target, each rock carries just a little energy.

Either way,

1. Energy of a rock is proportional to the size of the rock.
2. Intensity is defined as the energy that strikes the target per unit area per unit time.
3. Intensity is proportional to the number of rocks that strikes a unit area per unit time.

Note that it doesn't take very many blue rocks to achieve a given intensity. It takes a lot of red rocks.

But you don't see how many rocks or photons hit. You just see the intensity.

• Thank you for the comment! So if there's an equal number of blue rocks and red rocks that strike the target per unit area per unit time, then intensity would be greater in the wave of blue rocks? Or rather since we can't see the number of rocks, then if there's a wave with red rocks and a wave with blue rocks, and both waves have equal intensity, then we can infer that there are more red rocks that strike the target per unit area per unit time? Commented Jun 3, 2023 at 0:28
• Yes, exactly... Commented Jun 3, 2023 at 1:38

SI units here. We usually call $$f=photons/m^2/s$$ "photon flux", not "intensity". So, "intensity" $$I=W/m^2/s$$ is $$h \nu f$$. Thus, for a given intensity the flux is inversely proportional to frequency. This is why gamma ray astronomers have to coax information out of very few photons.

• Thank you! So photon flux = f = number of photons / (m^2 * s), and intensity = plank's constant * frequency of wave * photon flux. At a given intensity, higher frequency means less photon flux, which causes lower number of photons in gamma rays. However, if intensity is not a held constant, but rather photon flux is held constant, then a wave with greater frequency would have greater intensity? Commented Jun 3, 2023 at 0:25
• @syndromeofme Yes Commented Jun 3, 2023 at 11:11