Question

A radiation of energy 'E ' falls normally on a perfectly reflecting surface. The momentum transferred to the surface is (C = Velocity of light)

• A. $\frac{2E}{ C^2}$
• B. $ \frac{E}{C^2}$
• C. $\frac{E}{ C}$
• D. $\frac{2E}{ C}$

Answer 1

The Correct Option is D

To determine the momentum transferred to a perfectly reflecting surface when a radiation of energy 'E' falls normally on it, we need to understand the relationship between energy, momentum, and their interaction with a reflective surface.

When light (or radiation) of energy E strikes a perfectly reflective surface, it reflects back. The change in momentum provides an expression for how much momentum is transferred during this interaction.

Concept: The momentum p of radiation energy 'E' is given by the relation:

• p = \frac{E}{C}

where C is the speed of light.

For a perfectly reflecting surface, the radiation is reflected back. Therefore, the momentum change is double the initial momentum:

• Momentum transferred \Delta p = p_{\text{final}} - p_{\text{initial}} = \frac{E}{C} - \left(-\frac{E}{C}\right) = \frac{2E}{C}

Thus, the correct expression for the momentum transferred to the surface is \frac{2E}{C}.

Conclusion: Given the options, the correct answer is \frac{2E}{C}. This is because the perfectly reflecting surface causes the momentum to be doubled as it reflects the radiation back.