Question

Consider the situation described in the previous problem. Show that the force on the sphere due to the light falling on it is the same even if the sphere is not perfectly absorbing.

Answer 1

The force on the sphere due to the light falling on it is the same even if the sphere is not perfectly absorbing.

Explanation:

Step 1:

Find an O central sphere and an OP radius.

The sphere's radius OP makes an angle equal to OZ.

Let's rotate the radius around OZ to get the sphere into another circle.

The section of the sphere between the circle is an area ring $2 \pi r^{2} \sin \theta d \theta$.

Step 2:

Consider a small portion of the ring area ∆A at point P.

The energy of the light that falls in time ∆t on this section,

$\Delta U=I \Delta t(\Delta A \cos \theta)$

The change in momentum, as the light is reflected by the sphere along PR,

$\Delta p=\frac{2 \Delta U}{c} \cos \theta=\frac{2}{c} I \Delta t\left(\Delta A \cos ^{2} \theta\right)$

so,the force will be  

$\frac{\Delta p}{\Delta t}=\frac{2}{c} I \Delta A \cos ^{2} \theta$

The force part on A, alongside ZO, is

$\frac{\Delta p}{\Delta t} \cos \theta=\frac{2}{c} I \Delta A \cos ^{3} \theta$

Forcing intervention on the ring now,

$d F=\frac{2}{c} I\left(2 \pi r^{2} \sin \theta d \theta\right) \cos ^{3} \theta$

The force over the whole sphere,

$\begin{aligned}F &=\int_{0}^{\frac{\pi}{2}} \frac{4 \pi r^{2} I}{c} \cos ^{3} \theta \sin \theta d \theta \\=&-\int_{0}^{\frac{\pi}{2}} \frac{4 \pi r^{2} I}{c} \cos ^{3} \theta d(\cos \theta)=\frac{\pi r^{2} I}{c}\end{aligned}$