Basic Modern Physics
6.0 Radiation Pressure And Force
6.0 Radiation Pressure And Force
Intensity: The energy crossing per unit area per unit time perpendicular to the direction of the propagation is called the intensity of the light. So, $$\begin{equation} \begin{aligned} I = \frac{p}{A}\ W{m^{ - 2}} \\ \\\end{aligned} \end{equation} $$
Photon flux: It is defined as the number of photons incident on a normal surface per second per unit area. It is given by $$\phi = \frac{{Intensity}}{{Energy\ of\ photon}} = \frac{I}{{\left( {\frac{{hc}}{\lambda }} \right)}} = \frac{{I\lambda }}{{hc}}$$
Question 7. The sun delivers about $1.4\ kW{m^{ - 2}}$ of electromagnetic flux to the earth's surface. Calculate
(a) the total power on a roof of dimensions $8\ m \times 20\ m$
(b) the solar energy in joules incident on the roof in $1$ hour.
(c) the radiation pressure and the force assuming the roof to be a perfect absorber.
Solution: (a) As energy flux $I = \left[ {E/\left( {S \times t} \right)} \right] = \left[ {P/S} \right]$
$$\begin{equation} \begin{aligned} P = I \times S = \left( {1.4 \times {{10}^3}} \right) \times \left( {8 \times 20} \right) \\ \quad = 2.24 \times 105{\text{ W = 224 }}kW \\ \\\end{aligned} \end{equation} $$
(b) As power $P = \left( {E/t} \right)$
$$\begin{equation} \begin{aligned} E = P \times t = \left( {2.24 \times {{10}^5}} \right) \times \left( {60 \times 60} \right) \\ \quad = 8.064 \times {10^8} = 806.4\ MJ \\\end{aligned} \end{equation} $$
(c)For perfectly absorbing surface radiation pressure,
$$\begin{equation} \begin{aligned} \quad \quad P = \left( {F/S} \right) \\ \Rightarrow \quad F = P \times S = 4.7 \times {10^{ - 6}} \times \left( {8 \times 20} \right) \\ \quad \quad \quad \quad \quad \; = 7.52 \times {10^{ - 4}}{\text{ }}N \\\end{aligned} \end{equation} $$
