Light with an intensity of 1 kW/m² falls normally on a surface and is completely absorbed. The radiation pressure is: A) 1 kPa B) 3 * 10¹¹ Pa C) 1.7 * 10^-⁶ Pa D) 3.3 * 10^-⁶ Pa E) 6.7 * 10^-⁶ Pa

Question

Quizplus · Accepted Answer

Firstly, let's calculate the momentum of light using P = E/c, where E is the energy and c is the speed of light.

E = 1 kW/m² * 1 ms = 1 J/m² 
c = 3 × 10⁸ m/s

P = E/c = 1 J/m² / (3 × 10⁸ m/s) = 3.33 × 10⁻⁹ N•s/m²

Next, we can use the formula for radiation pressure:

p = 2P/ A 
where A is the area of the surface receiving the radiation. Since the light falls normally on the surface, we can assume the area is 1 m².

p = 2 * 3.33 × 10⁻⁹ N•s/m² / 1 m² = 6.67 × 10⁻⁹ N/m²

We can convert this to Pascals using the conversion factor 1 Pa = 1 N/m²:

p = 6.67 × 10⁻⁹ N/m² = 6.67 × 10⁻⁹ Pa

Finally, we can express the answer in scientific notation:

p = 6.67 × 10⁻⁹ Pa = 3.3 × 10¹⁰⁻⁹ Pa = 3.3 × 10¹ Pa

Therefore, the answer is D.

Light with an Intensity of 1 KW/m2 Falls Normally on a Surface

Light with an Intensity of 1 KW/m² Falls Normally on a Surface