Search any question & find its solution
Question:
Answered & Verified by Expert
An EM wave of intensity I falls on a surface kept in vacuum and exerts radiation pressure $p$ on it. Which of the following are true?
Options:
Solution:
2069 Upvotes
Verified Answer
The correct answers are:
Radiation pressure is $\frac{\mathrm{I}}{\mathrm{c}}$ if the wave is totally absorbed
,
Radiation pressure is $\frac{2 I}{\mathrm{c}}$ if the wave is totally reflected
,
Radiation pressure is in the range $\frac{\mathrm{I}}{\mathrm{c}} < \mathrm{p} < \frac{2 \mathrm{I}}{\mathrm{c}}$ for real surfaces
Radiation pressure is $\frac{\mathrm{I}}{\mathrm{c}}$ if the wave is totally absorbed
,
Radiation pressure is $\frac{2 I}{\mathrm{c}}$ if the wave is totally reflected
,
Radiation pressure is in the range $\frac{\mathrm{I}}{\mathrm{c}} < \mathrm{p} < \frac{2 \mathrm{I}}{\mathrm{c}}$ for real surfaces
As we know that, the radiation pressure $(\mathrm{P})$ is the force exerted by charge particle in electromagnetic wave on unit area of the surface, i.e., rate of change of momentum per unit area of the surface.
Then, the momentum per unit area per unit time
$$
=\frac{\text { Intensity }}{\text { Speed of wave }}=\frac{\mathrm{I}}{\mathrm{c}}
$$
The change in momentum per unit time per unit area $=$ $\frac{\Delta \mathrm{I}}{\mathrm{c}}=$ radiation pressure $\mathrm{p}=\frac{\Delta \mathrm{I}}{\mathrm{c}}$
Momentum of incident wave per unit area per second $=\frac{I}{c}$
When wave is fully absorbed by the surface, the momentum of the reflected wave per unit time per unit area $=0$.
Radiation pressure $(\mathrm{p})=$ change in momentum per unit area per second $=\frac{\Delta \mathrm{I}}{\mathrm{c}}=\frac{\mathrm{I}}{\mathrm{c}}-0=\frac{\mathrm{I}}{\mathrm{c}}$.
When wave is totally reflected, then momentum of the reflected wave per unit area per time $=-\frac{\mathrm{I}}{\mathrm{c}}$, Radiation pressure $\mathrm{p}=\frac{\mathrm{I}}{\mathrm{c}}-\left(-\frac{\mathrm{I}}{\mathrm{c}}\right)=\frac{2 \mathrm{I}}{\mathrm{c}}$. So variation of radiation pressure $\mathrm{P}$ lies between the range $\frac{\mathrm{I}}{\mathrm{c}} < \mathrm{p} < \frac{2 \mathrm{I}}{\mathrm{c}}$ for real surface.
Then, the momentum per unit area per unit time
$$
=\frac{\text { Intensity }}{\text { Speed of wave }}=\frac{\mathrm{I}}{\mathrm{c}}
$$
The change in momentum per unit time per unit area $=$ $\frac{\Delta \mathrm{I}}{\mathrm{c}}=$ radiation pressure $\mathrm{p}=\frac{\Delta \mathrm{I}}{\mathrm{c}}$
Momentum of incident wave per unit area per second $=\frac{I}{c}$
When wave is fully absorbed by the surface, the momentum of the reflected wave per unit time per unit area $=0$.
Radiation pressure $(\mathrm{p})=$ change in momentum per unit area per second $=\frac{\Delta \mathrm{I}}{\mathrm{c}}=\frac{\mathrm{I}}{\mathrm{c}}-0=\frac{\mathrm{I}}{\mathrm{c}}$.
When wave is totally reflected, then momentum of the reflected wave per unit area per time $=-\frac{\mathrm{I}}{\mathrm{c}}$, Radiation pressure $\mathrm{p}=\frac{\mathrm{I}}{\mathrm{c}}-\left(-\frac{\mathrm{I}}{\mathrm{c}}\right)=\frac{2 \mathrm{I}}{\mathrm{c}}$. So variation of radiation pressure $\mathrm{P}$ lies between the range $\frac{\mathrm{I}}{\mathrm{c}} < \mathrm{p} < \frac{2 \mathrm{I}}{\mathrm{c}}$ for real surface.
Looking for more such questions to practice?
Download the MARKS App - The ultimate prep app for IIT JEE & NEET with chapter-wise PYQs, revision notes, formula sheets, custom tests & much more.