Animated Solution for Physics - Dual Nature of Matter and Radiation: If $2.5\times 10^{-6}$ N average force is exerted by a light wave on a non-reflecting surface of $30{\text{ cm}}^2$ area during 40 min of time span, the energy flux of light just before it falls on the surface is ...... ${\text{W/cm}}^2$. (Round off to the nearest integer. Assume complete absorption and normal incidence conditions are there.)

Imagine a beam of light striking a completely black, non-reflecting surface. The surface absorbs all the incoming light, and in doing so, it experiences a physical push. This phenomenon is known as radiation pressure. We are given the average force exerted by this light, the area of the surface, and the time duration. Our goal is to find the energy flux, which is just another term for the intensity of the light.

Look closely at the physics here. Even though photons have no rest mass, they carry momentum. According to the de Broglie relation and Einstein's energy-momentum equation, the momentum $p$ of a photon is its energy $E$ divided by the speed of light $c$.

When light is completely absorbed by a non-reflecting surface, the entire momentum of the incident photons is transferred to the surface. By Newton's second law, force is the rate of change of momentum. Therefore, the total force $F$ exerted on the surface is the total energy incident per unit time divided by the speed of light.

Now, what exactly is energy flux? It is simply the intensity of light—the total energy falling per unit area per unit time. Let's denote energy flux as $I$.

From this definition, we can express the total energy as Intensity times Area times time ($E=IA\Delta t$). Let's substitute this energy back into our force equation.

Notice something beautiful? The time interval $\Delta t$ completely cancels out! The forty minutes given in the question is just a distractor designed to test your conceptual clarity. We get a clean master equation relating force, intensity, area, and the speed of light.

We need to find the energy flux, $I$. Rearranging our master equation, we get:

There is a massive catch here that traps many students. The area is given in square centimeters ($30{\text{ cm}}^2$), and the final answer is also requested in Watts per square centimeter (${\text{W/cm}}^2$).

If we keep the area as it is, the numerator ($F\times c$) gives Watts (since $\text{Newtons}\times \text{m/s}=\text{Joules/s}=\text{Watts}$), and the denominator gives square centimeters. No unit conversion is needed! Converting the area to square meters would just add unnecessary calculation steps and increase the risk of a silly mistake.

Let's substitute the values and get the answer.

Multiplying the numbers in the numerator: $2.5\times 3=7.5$. The powers of ten combine to give $10^{-6}\times 10^8=10^2=100$. So the numerator is $7.5\times 100=750$.

Divide that by thirty, and we get a perfect $25{\text{ W/cm}}^2$.

As a final thought experiment, what if the surface was a perfect mirror instead of being non-reflecting? The photons would bounce back, transferring double the momentum ($\Delta p=\frac{2E}{c}$). The force would be exactly twice as much! Always pay attention to the nature of the surface in such problems.