A beam of intensity Ireflects from a long, totally reflecting cylinder of radius R; the beam is perpendicular to the central axis of the cylinder and has a diameter largerthan$\mathbf{2}\mathit{R}$. What is the beam’s force per unit length on the cylinder?

• The intensity of the beam is$I$.
• The radius of the totally reflecting cylinder is.$R$
• The beam is perpendicular to the central axis of the cylinder.
• The diameter of the beam is larger than.$2R$

As the beam is perpendicular to the central axis, all the components of the force are canceled out. So, the total force is due to the components only. Using the pressure and the force relation of the electromagnetic wave, we can calculate the force per unit length.

Formulae:

The radiation pressure due to beam intensity,$p_r=\frac{2I\theta }{c}\cdot \cdot \cdot \cdot \cdot \cdot \left(1\right)$

The force due to the pressure,$dF=pdA\cdot \cdot \cdot \cdot \cdot \cdot \left(2\right)$

As the beam is perpendicular to the central axis, all they components of the force are cancelled out. So, the total force is due to thex components only.

The totalxcomponent force equation using can be written as:

$d{F}_x=2dFcos\theta \cdot \cdot \cdot \cdot \cdot \cdot \left(3\right)$

Where, the force is pressure per unit area and the pressure is radiation pressure.

So, the radiation pressure is given by the equation (2) in equation (3) as follows:

$d{F}_x=2\left(p_{r}dA\right)cos\theta \cdot \cdot \cdot \cdot \cdot \cdot \left(4\right)$

So, the force per unit length can be calculated integrating the above equation with substituting the pressure value of equation (1) as follows:

$\begin{array}{c}{F}_x=\int 2\left(\frac{2I\theta }{c}\right)cos\theta RLd\theta (\text{Areaofthesector,}dA=RLd\theta )\\ \frac{F_x}{L}=\int \left(\frac{4I\theta }{c}\right)cos\theta Rd\theta \\ =\underset{0}{\overset{\frac{\pi }{2}}{}}\frac{4I\theta }{c}cos\theta Rd\theta \\ =\frac{4IR}{c}\underset{0}{\overset{\frac{\pi }{2}}{}}\theta cos\theta d\theta \\ =\frac{4IR}{c}\underset{0}{\overset{\frac{\pi }{2}}{}}\theta cos\theta d\theta \\ =\frac{4IR}{c}\underset{0}{\overset{\frac{\pi }{2}}{}}\theta cos\theta d\theta \end{array}$

Substituting the given values$u=sin\theta ,du=cos\theta d\theta$in the above equation, we get the above value as:

$\begin{array}{c}\frac{F_x}{L}=\frac{4IR}{c}\underset{0}{\overset{\frac{\pi }{2}}{}}du\\ =\frac{4IR}{c}(u-\frac{u^3}{3})\\ =\frac{4IR}{c}{(sin\theta -\frac{(sin\theta ){}^3}{3})}_0^{\frac{\pi }{2}}\\ =\frac{4IR}{c}\{1-\frac{1}{3}\}\\ =\frac{4IR}{c}\left\{\frac{2}{3}\right\}\\ =\frac{8IR}{3c}\end{array}$

Hence, the value of beam’s force per length is.$\frac{8IR}{3c}$