A cone of radius $R$ and height $H$ with perfectly reflecting lateral surface, is placed in the path of a light beam of intensity I as shown. Calculate the force exerted by the light beam on this cone.

 

 

To find the force on cone, we consider an elemental strip of width dx on the lateral surface of cone at a distance $x$ from the vertex $O$ of cone as shown in figure.

If the radius of the strip is $r$, then surface area of the strip is

$dA=\left(2\pi r\right)dx$

Since for the cone, we have

$\sin \theta =\frac{r}{x}=\frac{R}{\sqrt{R^2+H^2}}$

$\Rightarrow r=x\sin \theta$

So, the area of strip is

$dA=2\pi \left(x\sin \theta \right)dx$

If $d{A}_n$ is the projection of area dA of the slant strip along the cross-sectional plane of the light beam i.e. normal to the light beam, then we have

$d{A}_n=dA\sin \theta$

If $dP$ is the power of light beam incident on this infinitesimal strip, then

$dP=Id{A}_n$
So, the force of on this infinitesimal strip element acts along the normal at the point of incidence and is given by

$df=\left(\frac{2dP}{c}\right)\sin \theta =\left(\frac{2Id{A}_n}{c}\right)\sin \theta =\frac{2IdA{\sin }^2\theta }{c}$

On resolving this infinitesimal force, we observe that the vertical components of the forces cancel, whereas the net force $F$ is only obtained by integrating the component $df\sin \theta$acting along the beam of light.

$F=\int df\sin \theta$

$\Rightarrow F=\int \left(\frac{2IdA}{c}{\sin }^2\theta \right)\sin \theta$

$\Rightarrow F=\int \left(\frac{2I}{c}{\sin }^3\theta \right)dA=\int \left(\frac{2I}{c}{\sin }^3\theta \right)\left[2\pi \left(x\sin \theta \right)dx\right]$

$\Rightarrow F=\left(\frac{4\pi I}{c}{\sin }^4\theta \right){\int }_0^{\sqrt{R^2+H^2}}xdx$

$\Rightarrow F=\left(\frac{4\pi l}{c}{\sin }^4\theta \right)\left({\left.\frac{x^2}{2}\right|}_0^{\sqrt{R^2+H^2}}\right)$

$\Rightarrow F=\left(\frac{2\pi I}{c}\right)\left[\frac{R^4}{{\left(R^2+H^2\right)}^2}\right]\left(R^2+H^2\right)$

$\Rightarrow F=\frac{2\pi I{R}^4}{c\left(R^2+H^2\right)}$