A sphere of radius 1.00 cm is positioned in the path of a parallel...

Let the intensity of the incident light be $I$ and the radius of the sphere be $R$.

For a perfectly absorbing sphere, the light is completely absorbed by the projected cross-sectional area $\pi R^2$. The force exerted is the rate of change of momentum:
$F_{abs} = \dfrac{I}{c} \pi R^2$

For a perfectly reflecting sphere, consider an elemental ring on the surface of the sphere at an angle $\theta$ with respect to the direction of the incident beam. The area of this element is $dS = 2\pi R^2 \sin \theta d\theta$.

The power incident on this elemental area is $P = I \cos \theta dS$.
Since the light is specularly reflected, the change in momentum per unit time (force) is directed along the normal to the surface:
$dF = \dfrac{2P}{c} \cos \theta = \dfrac{2I}{c} \cos^2 \theta dS$

The component of this force along the direction of the incident beam (z-axis) is:
$dF_z = dF \cos \theta = \dfrac{2I}{c} \cos^3 \theta dS$

Substituting $dS = 2\pi R^2 \sin \theta d\theta$, we get:
$dF_z = \dfrac{4\pi R^2 I}{c} \cos^3 \theta \sin \theta d\theta$

Integrating this over the illuminated hemisphere (from $\theta = 0$ to $\theta = \pi/2$):
$F_{ref} = \dfrac{4\pi R^2 I}{c} \int_0^{\pi/2} \cos^3 \theta \sin \theta d\theta$

Let $\cos \theta = t$, then $-\sin \theta d\theta = dt$. The limits change from $1$ to $0$:
$F_{ref} = \dfrac{4\pi R^2 I}{c} \int_0^1 t^3 dt = \dfrac{4\pi R^2 I}{c} \left[ \dfrac{t^4}{4} \right]_0^1 = \dfrac{\pi R^2 I}{c}$

We can see that $F_{ref} = F_{abs}$. The force on a perfectly reflecting sphere is exactly equal to the force on a perfectly absorbing sphere.

Since any opaque partially absorbing sphere can be considered as a combination of an absorbing fraction $\alpha$ and a reflecting fraction $\rho$ (where $\alpha + \rho = 1$), the total force will be:
$F = \alpha F_{abs} + \rho F_{ref} = (\alpha + \rho) \dfrac{\pi R^2 I}{c} = \dfrac{\pi R^2 I}{c}$

Therefore, the force on the sphere is the same as when it completely absorbs the radiation.