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Electromagnetic radiations of intensity I is incident normally on a perfectly reflecting surface. If speed of light is C, then radiation pressure exerted on the surface is
  • a)
    2I / C
  • b)
    I / C
  • c)
    2I / 3C
  • d)
    I / 2C
Correct answer is option 'A'. Can you explain this answer?
Verified Answer
Electromagnetic radiations of intensity I is incident normally on a pe...
Let no. of photons falling per second per unit area = n
So, I = nhf
Pressure = Force / Area = 2n(h / λ) = 2nhf / c = 2I / C
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Most Upvoted Answer
Electromagnetic radiations of intensity I is incident normally on a pe...
Radiation Pressure on a Perfectly Reflecting Surface

To find the radiation pressure exerted on a perfectly reflecting surface, we need to consider the properties of electromagnetic radiation and how it interacts with the surface.

Radiation Pressure:
Radiation pressure is the force exerted by electromagnetic radiation on a surface. When light or any other form of electromagnetic radiation strikes a surface, it imparts momentum to the surface, resulting in a force being exerted. The radiation pressure is directly proportional to the intensity of the incident radiation.

Perfectly Reflecting Surface:
A perfectly reflecting surface reflects all incident radiation without any absorption or transmission. This means that the surface does not absorb any radiation but reflects it completely.

Intensity of Incident Radiation:
The intensity of electromagnetic radiation is defined as the power per unit area. Mathematically, it is given by the equation:
Intensity (I) = Power (P) / Area (A)

Calculation of Radiation Pressure:
When electromagnetic radiation strikes a surface, it imparts momentum to the surface. The change in momentum per unit time is equal to the radiation pressure.

Let's consider a small area element (ΔA) on the surface. The change in momentum (Δp) of this area element due to the reflection of radiation is given by the equation:
Δp = 2IΔt / C

Here, Δt is the time taken for the radiation to reflect from the surface, and C is the speed of light.

Since the radiation is reflected normally, the change in momentum is in the opposite direction to the incident radiation. Therefore, the force exerted on the surface is given by:
F = Δp / Δt = 2I / C

Thus, the radiation pressure exerted on the surface is equal to 2I / C.

Answer:
Therefore, the correct answer is option 'A': 2I / C.
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Electromagnetic radiations of intensity I is incident normally on a perfectly reflecting surface. If speed of light is C, then radiation pressure exerted on the surface isa)2I / Cb)I / Cc)2I / 3Cd)I / 2CCorrect answer is option 'A'. Can you explain this answer?
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