Q.Two monochromatic beams A and B of equal intensity $I$, hit a screen. The number of photons hitting the screen by beam A is twice that by beam B. Then what inference can you make about their frequencies?

The key idea is frequency invariance: for monochromatic light of a given frequency, each photon carries a fixed energy $E = h\nu$, and the beam intensity is $I = n h \nu$, where $n$ is the photon flux (number per second per area).

Reasoning:

1. Given equal intensities: $I_A = I_B$.
2. Intensity relation: $n_A h \nu_A = n_B h \nu_B$. …

The key here is frequency invariance of a different kind: intensity is the same, but the photon numbers differ. That forces the photon energies — and therefore the frequencies — to adjust.

Let’s unpack why.

1. What does “intensity” mean for a monochromatic beam?

Intensity $I$ is the power per unit area. For a beam of light, this power comes from the energy carried by individual photons. Each photon of frequency $\nu$ carries energy $E = h\nu$, where $h$ is Planck’s constant.

If $n$ photons hit a unit area per second (the photon flux), then the intensity is:

This is the fundamental relation: intensity = (photon flux) × (energy per photon).

2. Set up the given conditions

Let beam A have frequency $\nu_A$ and photon flux $n_A$ (photons per second per unit area).

Let beam B have frequency $\nu_B$ and photon flux $n_B$.

We are told:

• Equal intensity: $I_A = I_B$
• Beam A has twice the photon count of beam B: $n_A = 2 n_B$

3. Write the intensity equality

From the relation above:

Since $I_A = I_B$:

Cancel $h$ (it’s the same constant):

4. Substitute the photon flux relation

We know $n_A = 2 n_B$. Plug that in:

Cancel $n_B$ (non-zero):

So:

$\nu_A = \frac{\nu_B}{2}$ …

Method: Relating Intensity, Photon Flux, and Frequency

Whenever two beams are compared through their intensity (or power) together with their photon counts, the bridge between the two is the relation $I = n\,h\nu$ — intensity equals photon flux times energy per photon.

Steps

Step 1: Write the intensity relation for each beam

where $n$ is the number of photons per second (per unit area), and $\nu$ is the frequency, so $h\nu$ is the energy per photon.

Step 2: Use the given equality/ratio of intensities to link the two beams

If the intensities are equal, set $n_A h \nu_A = n_B h \nu_B$; the constant $h$ cancels immediately, leaving $n_A \nu_A = n_B \nu_B$. …