50 W/m² energy density of sunlight is normally incident on the surface of a solar panel. Some part o

Q: Why is the pressure for reflection double that of absorption?

It comes down to the conservation of momentum. When a photon is absorbed, its momentum goes from $p$ to $0$, so the change is $p$. When a photon is reflected, it bounces back, going from $p$ to $-p$. The change in momentum is $p - (-p) = 2p$. Since force is the rate of change of momentum, the force (and thus pressure) is doubled for reflection.

Q: Why do we use intensity 'I' instead of energy in the formula?

Intensity $I$ is defined as Power per unit Area, which is Energy per unit Time per unit Area. Radiation pressure is Force per unit Area, which is Momentum per unit Time per unit Area. Because Energy and Momentum of a photon are related by $E = pc$, Intensity and Pressure are directly related by $I = Pc$.

Q: Is the force really this small in real life? Why don't we feel it?

Yes, radiation pressure is extremely weak. A force of $20 \times 10^{-8}$ Newtons is microscopic. For comparison, a small apple resting on your hand exerts a downward force of about $1$ Newton. The push of sunlight is completely overwhelmed by gravity, wind, and our own muscle movements.

Q: What if the light was incident at an angle instead of normally?

If the light hits at an angle $\theta$ to the normal, the momentum change perpendicular to the surface is reduced by a factor of $\cos\theta$. Additionally, the beam spreads out over a larger area, reducing the effective intensity by another factor of $\cos\theta$. So, the pressure would scale by $\cos^2\theta$.

Q: Does the area of the panel affect the radiation pressure?

No, radiation pressure is independent of the area, just like atmospheric pressure. However, the total *force* exerted on the panel does depend on the area, because Force = Pressure $\times$ Area.

Question

50 W/m$^2$ energy density of sunlight is normally incident on the surface of a solar panel. Some part of incident energy (25%) is reflected from the surface and the rest is absorbed. The force exerted on 1 m$^2$ surface area will be close to ($c = 3 	imes 10^8$ m/s)

Options:

(A) $20 	imes 10^{-8}$ N

(B) $35 	imes 10^{-8}$ N

(C) $15 	imes 10^{-8}$ N

(D) $10 	imes 10^{-8}$ N

Sigma Percentile · Accepted Answer

### The Hidden Push of Sunlight

Have you ever stood outside on a sunny day and felt the warmth of the sun on your skin? While you can easily feel the heat, there is something else happening that is completely invisible to our senses: the sunlight is actually *pushing* you.

Light carries momentum. Even though photons have no mass, they possess momentum given by the equation $p = \frac{E}{c}$, where $E$ is the energy and $c$ is the speed of light. When billions of these photons strike a surface, they transfer their momentum, creating a continuous force. We call this phenomenon **radiation pressure**.

### The Physics of Bouncing vs. Absorbing

To solve this problem, we need to understand how light interacts with a surface. There are two extreme cases:

1. **Perfect Absorption (A Black Body):** If a surface absorbs all the light, the photons hit it and stop. The change in momentum for each photon is simply its initial momentum, $p$. The resulting radiation pressure is $P = \frac{I}{c}$, where $I$ is the intensity of the light.

2. **Perfect Reflection (A Mirror):** If a surface reflects all the light, the photons bounce back with the same speed but in the opposite direction. The change in momentum is $p - (-p) = 2p$. Because the momentum change is doubled, the radiation pressure is also doubled: $P = \frac{2I}{c}$.

### Analyzing the Solar Panel

Our solar panel is neither a perfect mirror nor a perfect black body. The problem states that it reflects $25\%$ of the incident light and absorbs the remaining $75\%$.

Because pressure is a linear physical quantity, we can simply take a weighted average of the two extreme cases. The total radiation pressure $P_{total}$ will be the sum of the pressure from the reflected part and the pressure from the absorbed part:

$$P_{total} = 0.25 \left( \frac{2I}{c} ight) + 0.75 \left( \frac{I}{c} ight)$$

Let's convert those decimals into fractions to make the algebra cleaner. $25\%$ is $\frac{1}{4}$ and $75\%$ is $\frac{3}{4}$.

$$P_{total} = \frac{1}{4} \left( \frac{2I}{c} ight) + \frac{3}{4} \left( \frac{I}{c} ight)$$

$$P_{total} = \frac{1}{2} \frac{I}{c} + \frac{3}{4} \frac{I}{c}$$

By finding a common denominator, we get:

$$P_{total} = \frac{2}{4} \frac{I}{c} + \frac{3}{4} \frac{I}{c} = \frac{5}{4} \frac{I}{c}$$

### Calculating the Final Force

We are asked to find the **force** exerted on a $1 	ext{ m}^2$ surface area. We know that pressure is defined as force per unit area ($P = \frac{F}{A}$). Therefore, the force is $F = P_{total} 	imes A$.

Substituting our expression for total pressure:

$$F = \frac{5}{4} \frac{I}{c} 	imes A$$

Now, we plug in the given values: the intensity $I = 50 	ext{ W/m}^2$, the speed of light $c = 3 	imes 10^8 	ext{ m/s}$, and the area $A = 1 	ext{ m}^2$.

$$F = \frac{5}{4} 	imes \frac{50}{3 	imes 10^8} 	imes 1$$

$$F = \frac{250}{12 	imes 10^8}$$

$$F \approx 20.83 	imes 10^{-8} 	ext{ N}$$

Looking at our options, the closest value is **$20 	imes 10^{-8} 	ext{ N}$**.

While this force is incredibly small—roughly the weight of a grain of sand—it is constant and requires no fuel. This exact principle is what engineers use to design **solar sails**, allowing spacecraft to accelerate continuously through the vacuum of space using nothing but the gentle push of starlight!

Select Answer:

Visualized Solution

Visualizing the Setup

Radiation Pressure Formulas

Analyzing the Panel's Properties

Setting up the Total Pressure Equation

Simplifying the Expression

Relating Pressure to Force

Substituting the Values

Final Calculation

The Sigma Insight: Characteristics of Electromagnetic Waves

The Hidden Push of Sunlight

The Physics of Bouncing vs

Analyzing the Solar Panel

Calculating the Final Force

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