Question

Light with an energy flux of $ 25 \times 10^4 W m^{-2} $ falls on a perfectly reflecting surface at normal incidence. If the surface area is $ 15 c m^2 $, the average force exerted on the surface is

• A. $1.25 \times 10^{-6} N $
• B. $2.50\times10^{-6} N$
• C. $1.20\times10^{-6}N$
• D. $3.0\times10^{-6}N$

Answer 1

The Correct Option is B

To find the average force exerted on the surface, we must understand how light exerts pressure. When light reflects perfectly off a surface, it transfers momentum. The force can be calculated using the following relationships:

• Energy flux (or intensity), denoted as I, is given by I = 25 \times 10^{4} \, \text{W/m}^2.
• The area A of the surface is 15 \, \text{cm}^2, which must be converted to \text{m}^2: A = 15 \, \text{cm}^2 = 15 \times 10^{-4} \, \text{m}^2.

The pressure P exerted by the light on the surface is given by P = \frac{2I}{c}, where c is the speed of light, approximately 3 \times 10^{8} \, \text{m/s}.

1. Calculate the pressure:

   P = \frac{2 \times 25 \times 10^{4}}{3 \times 10^{8}}

   P = \frac{50 \times 10^{4}}{3 \times 10^{8}}

   P = \frac{50}{3} \times 10^{-4} \, \text{N/m}^2

   P = 16.67 \times 10^{-4} \, \text{N/m}^2

2. Calculate the force F using the formula F = P \times A:

   F = 16.67 \times 10^{-4} \times 15 \times 10^{-4} \, \text{N}

   F = 250.05 \times 10^{-8} \, \text{N}

   F \approx 2.50 \times 10^{-6} \, \text{N}

The average force exerted on the surface is therefore 2.50\times10^{-6} \, \text{N}, which matches the given correct answer option:

2.50\times10^{-6} N