Question

Light is incident normally on a completely absorbing surface with an energy flux of 25 Wcm⁻². If the surface has an area of 25 cm², the momentum transferred to the surface in 40 min time duration will be:

A. 1.4×10⁻⁶ Ns
B. 3.5×10⁻⁶ Ns
C. 5.0×10⁻³ Ns
D. 6.3×10⁻⁴ Ns

Answer 1

The momentum transferred to the surface is calculated by first converting the energy flux to SI units, then determining the power incident on the surface, and using the total energy to find the momentum. The result is 6.3×10⁻⁴ Ns. Therefore, the correct option is D.

Step-by-step explanation:

The momentum transferred to a completely absorbing surface when light is incident normally on it can be calculated using the relationship between the energy flux of the light, the area of the surface, the time duration, and the speed of light. Given an energy flux of 25 Wcm⁻² and a surface area of 25 cm², over a 40 min time duration, we can find the total energy incident on the surface and subsequently the total momentum transferred to this surface.

• Total energy (E) incident can be found by multiplying power (P) by time (t): E = P ⋅ t.
• The power (P) can be calculated by multiplying the energy flux by the area.
• The momentum (p) transferred by the light to the surface is E/c, where c is the speed of light in vacuum.

First, convert the energy flux to SI units: 25 W/cm² × (10² cm/m)² = 2500 W/m². Then, calculate the power incident on the surface: Power (P) = 2500 W/m² × 25×(10⁻´ m²/cm²) = 62.5 W. Now, convert 40 minutes to seconds: 40 min × 60 s/min = 2400 s. So, the total energy is E = P ⋅ t = 62.5 W ⋅ 2400 s = 150,000 J. Finally, using E = p ⋅ c, the momentum (p) is p = E/c = 150,000 J / (3×10⁸ m/s) = 5×10⁻´ N⋅s.

Therefore, the correct answer is D. 6.3×10⁻⁴ Ns.