Let's analyze the effect of decreasing the area of a totally absorbing surface on radiation pressure when light of uniform intensity shines perpendicularly on it.

Understanding Radiation Pressure

Radiation pressure is the pressure exerted upon any surface exposed to electromagnetic radiation. It is essentially the momentum transferred by photons to the surface per unit area per unit time.

Radiation Pressure on a Totally Absorbing Surface

When light of intensity \(I\) shines perpendicularly on a totally absorbing surface, the radiation pressure \(P_r\) is given by the formula:

\begin{equation*} P_r = \frac{I}{c} \end{equation*}

where:

• \(I\) is the intensity of the light.
• \(c\) is the speed of light in vacuum.

This formula tells us that the radiation pressure depends directly on the intensity of the light and inversely on the speed of light.

Analyzing the Given Scenario

The question specifies several conditions:

• Light has uniform intensity (\(I\)). This means the intensity \(I\) is constant across the illuminated surface and does not change.
• The surface is totally absorbing. This is important because the formula \(P_r = \frac{I}{c}\) is specifically for totally absorbing surfaces under perpendicular incidence.
• Light shines perpendicularly on the surface. This confirms the applicability of the simple \(P_r = \frac{I}{c}\) formula without needing to consider angles.
• The surface is fully illuminated. This means the uniform intensity \(I\) is present over the entire area being considered.

The question asks what happens to the radiation pressure if the area of the surface is decreased.

Effect of Decreasing Area on Radiation Pressure

Looking at the formula for radiation pressure, \(P_r = \frac{I}{c}\):

• The intensity \(I\) is given as uniform and does not change according to the problem statement.
• The speed of light \(c\) is a fundamental constant and does not change.

The formula \(P_r = \frac{I}{c}\) does not include the area of the surface \(A\). The radiation pressure is an intrinsic property of the incident radiation (its intensity) and the surface's interaction with it (total absorption), independent of the size of the surface being illuminated, as long as the intensity is uniform over that area.

While the total radiation force (\(F = P_r \times A\)) on the surface would decrease if the area \(A\) decreases (since \(P_r\) remains constant), the pressure itself (force per unit area) remains unchanged because it is determined solely by the intensity of the incident light and the speed of light in this specific scenario.

Therefore, if the area of the totally absorbing surface is decreased, the radiation pressure on the surface remains the same, provided the intensity of the light and its perpendicular incidence are maintained.

Summary

The radiation pressure on a totally absorbing surface due to perpendicularly incident light of uniform intensity is determined by the intensity of the light and the speed of light. It is independent of the area of the surface. Decreasing the area will decrease the total force, but not the pressure.

Revision Table: Radiation Pressure Concepts

Additional Information on Radiation Pressure

• For a perfectly reflecting surface with perpendicular incidence, the radiation pressure is \(P_r = \frac{2I}{c}\) because the momentum change is doubled upon reflection.
• For a surface with reflectivity \(R\) and absorptivity \(A\) (where \(R+A=1\)) at perpendicular incidence, the pressure is \(P_r = (1+R)\frac{I}{c} = (2-A)\frac{I}{c}\). For a totally absorbing surface, \(A=1, R=0\), giving \(P_r = \frac{I}{c}\).
• Radiation pressure is responsible for phenomena like the tails of comets pointing away from the sun (due to pressure on dust particles) and is used in technologies like optical tweezers.