Sunlight of intensity 50 W m$$^{-2}$$ is incident normally 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 ($$c = 3 \times 10^8$$ m s$$^{-1}$$):

Step-by-Step Solution

1. Identify Given Data

• Intensity of sunlight ($$I$$) = $$50\text{ W m}^{-2}$$
• Area of the solar panel ($$A$$) = $$1\text{ m}^2$$
• Speed of light ($$c$$) = $$3 \times 10^8\text{ m s}^{-1}$$
• Percentage of light reflected ($$a_r$$) = $$25\%$$
• Percentage of light absorbed ($$a_a$$) = $$75\%$$
• Force due to complete absorption:
• Force due to complete reflection (normally incident):

2. Formulate the Radiation Force Equation

The force exerted by electromagnetic radiation on a surface consists of two contributions: the force due to the absorbed light and the force due to the reflected light.

$$F_{\text{absorbed}} = \frac{I A}{c}$$

$$F_{\text{reflected}} = \frac{2 I A}{c}$$

Since $$25\%$$ of the light is reflected and $$75\%$$ is absorbed, the total force ($$F$$) acting on the area is:

$$F = \frac{25}{100} \left( \frac{2 I A}{c} \right) + \frac{75}{100} \left( \frac{I A}{c} \right)$$

3. Simplify the Expression

Substitute $$A = 1\text{ m}^2$$:

$$F = \frac{50}{100} \left( \frac{I}{c} \right) + \frac{75}{100} \left( \frac{I}{c} \right)$$

$$F = \frac{125}{100} \left( \frac{I}{c} \right)$$

4. Substitute Values and Calculate

Substitute the values of $$I$$ and $$c$$:

$$F = \frac{125}{100} \times \frac{50}{3 \times 10^8}$$

$$F = \frac{125 \times 0.5}{3 \times 10^8}$$

$$F = \frac{62.5}{3} \times 10^{-8}$$

$$F \approx 20.83 \times 10^{-8}\text{ N}$$

Final Answer

The force exerted on the surface is close to $$20.8 \times 10^{-8}\text{ N}$$.