Problem

Problem

Light with an energy flux of 20 W/cm220 \text{ W/cm}^2 falls on a non-reflecting surface at normal incidence. If the surface has an area of 30 cm230 \text{ cm}^2. The total momentum delivered (for complete absorption) during 30 minutes is a. 36×10−5 kg⋅m/s36 \times 10^{-5} \text{ kg}\cdot\text{m/s} b. 36×10−4 kg⋅m/s36 \times 10^{-4} \text{ kg}\cdot\text{m/s} c. 1.08×104 kg⋅m/s1.08 \times 10^4 \text{ kg}\cdot\text{m/s} d. 1.08×107 kg⋅m/s1.08 \times 10^7 \text{ kg}\cdot\text{m/s}
Intuitive explanation

When light is absorbed by a surface, it transfers momentum—not just energy. The key insight is that photons carry momentum p=E/cp = E/c, so the momentum flux (momentum per unit area per unit time) equals the energy flux divided by cc. Multiply by the surface area and duration to get total momentum delivered. This is why even "weightless" light can exert measurable force on matter.

Intuitive explanation

When light is absorbed by a surface, it transfers momentum—not just energy. The key insight is that photons carry momentum p=E/cp = E/c, so the momentum flux (momentum per unit area per unit time) equals the energy flux divided by cc. Multiply by the surface area and duration to get total momentum delivered. This is why even "weightless" light can exert measurable force on matter.

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