Light of uniform intensity shines perpendicularly on a totally absorbing surface, fully illuminating the surface. If the area of the surface is decreased, what is the effect on radiation pressure?
Remains the same
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.
| Factor | Change | Effect on \(P_r = \frac{I}{c}\) |
|---|---|---|
| Intensity (\(I\)) | Remains Uniform (Constant) | No change |
| Speed of Light (\(c\)) | Constant | No change |
| Area (\(A\)) | Decreased | No direct effect on \(P_r\) formula |
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
| Concept | Description | Formula (Totally Absorbing, Perpendicular) |
|---|---|---|
| Radiation Intensity (\(I\)) | Power per unit area | \(I = P/A\) |
| Radiation Pressure (\(P_r\)) | Force per unit area exerted by light | \(P_r = I/c\) |
| Radiation Force (\(F\)) | Total force exerted by light | \(F = P_r \times A\) |
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.
Important Questions from Ray Optics and Optical Instruments
A Convex mirror produces the magnification 1/3 and 1/4 when the object is placed at the points P and Q in front of the mirror.
Which of the following statements are correct?
- A. The saturation current is constant with collector plate potential for different frequencies of incident radiation.
- B. The saturation current is different with collector plate potential for different frequencies of incident radiation.
- C. The saturation current is different with collector plate potential for different intensity of incident radiation.
- D. The saturation current is constant with collector plate potential for different intensity of incident radiation.
- E. Below threshold frequency, no photoelectrons are emitted.
Choose the correct answer from the options given below:
For insulators and semiconductors, the resistance decreases with an increase in temperature because:
A ray of light passes through four transparent media with refractive index μ1, μ2, μ3, and μ4 as shown in the figure. The surfaces of all media are parallel. If BC and DE are parallel, we must have:
A ray of light passes through an equilateral glass prism in such a manner that the angle of incidence is equal to the angle of emergence, and each of these angles is equal to (3/4)th of the angle of the prism. The angle of deviation is:
