Question 14 Marks
An electromagnetic wave transports linear momentum as it travels through space. If an electromagnetic wave transfers a total energy $U$ to a surface in time $t,$ then total linear momentum delivered to the surface is $\text{p}=\frac{\text{U}}{\text{c}}.$ When an electromagnetic wave falls on a surface, it exerts pressure on the surface. ln $1903,$ the American scientists Nichols and Hull succeeded in measuring radiation pressures of visible light where other had failed, by making a detailed empirical analysis of the ubiquitous gas heating and ballistic effects.
- The pressure exerted by an electromagnetic wave of intensity $I (Wm^{-2})$ on a non$-$reflecting surface is $(c$ is the velocity of light$).$
- $\text{Ic}$
- $\text{Ic}^2$
- $\frac{\text{I}}{\text{c}}$
- $\frac{\text{I}}{\text{c}^2}$
- Light with an energy flux of $18\frac{\text{W}}{\text{cm}^2}$ falls on a non$-$reflecting surface at normal incidence. The pressure exerted on the surface is:
- $3\frac{\text{N}}{\text{m}^2}$
- $2\times10^{-4}\frac{\text{N}}{\text{m}^2}$
- $6\frac{\text{N}}{\text{m}^2}$
- $6\times10^{-4}\frac{\text{N}}{\text{m}^2}$
- Radiation of intensity $0.5Wm^{-2}$ are striking a metal plate. The pressure on the plate is:
- $0.166 \times 10^{-8}Nm^{-2}$
- $0.212 \times 10^{-8}Nm^{-2}$
- $0.132 \times 10^{-8}Nm^{-2}$
- $0.083 \times 10^{-8}Nm^{-2}$
- A point source of electromagnetic radiation has an average power out$-$put of $1500W.$ The maximum value of electric field at a distance of 3m from this source $($in $Vm^{-1})$ is:
- $500$
- $100$
- $\frac{500}{3}$
- $\frac{250}{3}$
- The radiation pressure of the visible light is of the order of,
- $10^{-2}\frac{\text{N}}{\text{m}^2}$
- $10^{-4}\frac{\text{N}}{\text{m}}$
- $10^{-6}\frac{\text{N}}{\text{m}^2}$
- $10^{-8}\text{N}$
Answer
$=\frac{18\times10^4\frac{\text{W}}{\text{cm}^2}}{3\times8^8\frac{\text{m}}{\text{s}}}=6\times10^{-4}\frac{\text{N}}{\text{m}^2}$
$\text{V}_\text{av}=\frac{1}{2}\epsilon_0\text{E}_0^2\times\text{c}$
$\Rightarrow\text{E}_0=\sqrt{\frac{\text{P}}{2\pi\text{R}^2\epsilon_0\text{c}}}=\sqrt{\frac{1500}{2\times3.14(3)^2\times8.85\times10^{-12}\times3\times18^8}}$
$=\sqrt{10,000}=100\text{Vm}^{-1}$
View full question & answer→- $(c) \frac{\text{I}}{\text{c}}$
- $(d) 6\times10^{-4}\frac{\text{N}}{\text{m}^2}$
$=\frac{18\times10^4\frac{\text{W}}{\text{cm}^2}}{3\times8^8\frac{\text{m}}{\text{s}}}=6\times10^{-4}\frac{\text{N}}{\text{m}^2}$
- $(a)\ 0.166 \times 10^{-8}Nm^{-2}$
- $(b)\ 100$
$\text{V}_\text{av}=\frac{1}{2}\epsilon_0\text{E}_0^2\times\text{c}$
$\Rightarrow\text{E}_0=\sqrt{\frac{\text{P}}{2\pi\text{R}^2\epsilon_0\text{c}}}=\sqrt{\frac{1500}{2\times3.14(3)^2\times8.85\times10^{-12}\times3\times18^8}}$
$=\sqrt{10,000}=100\text{Vm}^{-1}$
- $(c) 10^{-6}\frac{\text{N}}{\text{m}^2}$
