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Light with an energy flux of \(18\ \text{W/cm}^2\)  falls on a non reflecting surface at normal incidence. If the surface has an area of \(20\;\text c\text m^2\)  find the average force exerted on the surface during a \(30\) minute time span.

Answer

\(\text{Energy}=18\ \text{W/cm}^2\)
\(\text{Area}=20\;\text c\text m^2\)
\(\text{Time}=30\) minutes
\(\text{c}=\) Speed of light \(=3\times 10^8\ \text{m/s}\)
\(\text{Power}=\frac{\text{Energy}}{\text{time}}\)
As we have given power/area
The total energy \(=\)  energy \(\times \)  area \(\times \)  time  
Total energy
\(\text{U}=18\times 20\times 30\times 60\)
\(\text{U}=6.48\times 10^5\text J\)
The momentum transferred when they strike the surface will be \(\text{U/c}\)
\(p=\frac{\text{U}}{\text{c}}\)
\(p=\frac{6.48\times 10^5}{3\times 10^8}\)
\(p=2.16\times 10^{-3}\;\text{kg/m}\)
Therefore the average force exerted will be
\(F=\frac p t\)
Here \(t\)  is the time period \(30\text{ min}=1800\text{ sec}\)
\(F=\frac{2.16\times 10^{-3}}{1800}\)
\(F=1.2\times 10^{-6}\;\text{N}\)
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