Question

Light with an energy flux$18W{m}^{-3}$ falls on a non-reflecting surface at normal to the surface. If the surface has an area of$20m^2$. The average force exerted on the surface $30minutes$ is

• A. $6·48x{10}^{5}N$
• B. $3·60x{10}^{2}N$
• C. $1·2x{10}^{-6}N$
• D. $2·16x{10}^{-3}N$

Answer 1

The correct option is C

$1·2x{10}^{-6}N$

Step 1. Given data:

1. The energy flux of light or intensity$I=18W{m}^{-3}$
2. Area of the surface$A=20m^2$
3. The time duration$∆t=30minutes=30\times 60s=1800s$

Step 2. Formula used:

1. Total energy falling on the surface,
   $U=IA∆t$ ($I=$Energy flux, $A=$Area of the surface, $∆t=$Time duration)
2. The momentum produced,
   $p=\frac{U}{c}$ ($c=$speed of light, $U=$Total energy falling on the surface)
3. The average force exerted on the surface,
   $F=\frac{p}{∆t}$ (symbols have their usual meaning)

Step 3. Calculating total energy:

Total energy falling on the surface,

$\begin{array}{rcl}U& =& IA∆t\\ & =& 18\times 20\times 1800J\\ & =& 6·48\times {10}^{5}J\end{array}$

Step 3. Calculating the momentum:

The momentum produced,

$\begin{array}{rcl}p& =& \frac{U}{c}\\ & =& \frac{6·48\times {10}^5}{3\times {10}^8}kgm{s}^{-1}\\ & =& 2·16\times {10}^{-3}kgm{s}^{-1}\end{array}$ (symbols have their usual meaning)

Step 4. Calculating the average force:

The average force exerted on the surface,

$\begin{array}{rcl}F& =& \frac{p}{∆t}\\ & =& \frac{2·16\times {10}^{-3}}{1800}N\\ & =& 1·2\times {10}^{-6}N\end{array}$ (symbols have their usual meaning)

Thus, the average force exerted on the surface$1·2\times {10}^{-6}N$.

Hence, option C is the correct answer.