Question
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.
  1. The pressure exerted by an electromagnetic wave of intensity I $(Wm^{-2})$ on a non-reflecting surface is (c is the velocity of light).
  1. $\text{Ic}$
  2. $\text{Ic}^2$
  3. $\frac{\text{I}}{\text{c}}$
  4. $\frac{\text{I}}{\text{c}^2}$
  1. 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:
  1. $3\frac{\text{N}}{\text{m}^2}$
  2. $2\times10^{-4}\frac{\text{N}}{\text{m}^2}$
  3. $6\frac{\text{N}}{\text{m}^2}$
  4. $6\times10^{-4}\frac{\text{N}}{\text{m}^2}$
  1. Radiation of intensity $0.5Wm^{-2}$ are striking a metal plate. The pressure on the plate is:
  1. $0.166 \times 10^{-8}Nm^{-2}$
  2. $0.212 \times 10^{-8}Nm^{-2}$
  3. $0.132 \times 10^{-8}Nm^{-2}$
  4. $0.083 \times 10^{-8}Nm^{-2}$
  1. 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:
  1. $500$
  2. $100$
  3. $\frac{500}{3}$
  4. $\frac{250}{3}$
  1. The radiation pressure of the visible light is of the order of,
  1. $10^{-2}\frac{\text{N}}{\text{m}^2}$
  2. $10^{-4}\frac{\text{N}}{\text{m}}$
  3. $10^{-6}\frac{\text{N}}{\text{m}^2}$
  4. $10^{-8}\text{N}$

Answer

  1. (c) $\frac{\text{I}}{\text{c}}$
Explanation:
Pressure exerted by an electromagnetic radiation, $\text{P}=\frac{\text{I}}{\text{c}}.$
  1. (d) $6\times10^{-4}\frac{\text{N}}{\text{m}^2}$
Explanation:
$\text{p}_\text{rad}=\frac{\text{Energy flux}}{\text{Speed of light}}=\frac{18\frac{\text{W}}{\text{cm}^2}}{3\times8^8\frac{\text{m}}{\text{s}}}$
$=\frac{18\times10^4\frac{\text{W}}{\text{cm}^2}}{3\times8^8\frac{\text{m}}{\text{s}}}=6\times10^{-4}\frac{\text{N}}{\text{m}^2}$
  1. (a) $0.166 \times 10^{-8}Nm^{-2}$
Explanation:
$\text{p}=\frac{\text{I}}{\text{c}}=\frac{0.5}{3\times10^8}=0.166\times10^{-8}\text{Nm}^{-2}$
  1. (b) $100$
Explanation:
Intensity of EM wave is given by $\text{I}=\frac{\text{P}}{4\pi\text{R}^2}$
$\text{V}_\text{av}=\frac{1}{2}\in_0\text{E}_0^2\times\text{c}$
$\Rightarrow\text{E}_0=\sqrt{\frac{\text{P}}{2\pi\text{R}^2\in_0\text{c}}}=\sqrt{\frac{1500}{2\times3.14(3)^2\times8.85\times10^{-12}\times3\times18^8}}$
$=\sqrt{10,000}=100\text{Vm}^{-1}$
  1. (c) $10^{-6}\frac{\text{N}}{\text{m}^2}$
Explanation:
The radiation pressure of visible light $=7\times10^{-6}\frac{\text{N}}{\text{m}^2}$

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