1. Concept and Setup

The force exerted on the lens is equal to the rate of change of momentum of the photons interacting with it. The lens changes the direction of the incoming diverging beam into a parallel outgoing beam.

$$ \vec{F}_{lens} = \frac{d\vec{p}_{out}}{dt} – \frac{d\vec{p}_{in}}{dt} $$

Let $\theta_0$ be the semi-vertical angle subtended by the lens at the source $S$. From the geometry:

$$ \sin\theta_0 = \frac{r}{\sqrt{r^2 + d^2}}, \quad \cos\theta_0 = \frac{d}{\sqrt{r^2 + d^2}} $$

2. Incoming Momentum Flux

The source is isotropic with power $P$. The power per unit solid angle is $P/4\pi$. The momentum of a photon is $E/c$. We need the component of momentum along the optical axis (z-axis).

$$ F_{in} = \int_{0}^{\theta_0} \left( \frac{dP}{c} \right) \cos\theta $$

$$ F_{in} = \int_{0}^{\theta_0} \frac{1}{c} \left( \frac{P}{4\pi} 2\pi \sin\theta d\theta \right) \cos\theta = \frac{P}{2c} \int_{0}^{\theta_0} \sin\theta \cos\theta d\theta $$

$$ F_{in} = \frac{P}{2c} \left[ \frac{\sin^2\theta}{2} \right]_0^{\theta_0} = \frac{P}{4c} \sin^2\theta_0 $$

3. Outgoing Momentum Flux

The lens converts the intercepted light into a beam parallel to the axis. The total power intercepted by the lens depends on the solid angle $\Omega = 2\pi(1-\cos\theta_0)$.

$$ P_{intercepted} = \frac{P}{4\pi} \times 2\pi(1-\cos\theta_0) = \frac{P}{2}(1-\cos\theta_0) $$

Since the outgoing beam is parallel to the axis, the full momentum contributes to the force:

$$ F_{out} = \frac{P_{intercepted}}{c} = \frac{P}{2c}(1-\cos\theta_0) $$

4. Net Force Calculation

The force on the lens is the difference between the momentum flux leaving the lens and the momentum flux entering it.

$$ F_{net} = F_{out} – F_{in} = \frac{P}{2c}(1-\cos\theta_0) – \frac{P}{4c}\sin^2\theta_0 $$

Using the identity $\sin^2\theta_0 = 1 – \cos^2\theta_0$:

$$ F_{net} = \frac{P}{2c} \left[ (1-\cos\theta_0) – \frac{1}{2}(1-\cos^2\theta_0) \right] $$

$$ F_{net} = \frac{P}{2c} \left[ 1 – \cos\theta_0 – \frac{1}{2} + \frac{1}{2}\cos^2\theta_0 \right] = \frac{P}{2c} \left[ \frac{1}{2} – \cos\theta_0 + \frac{1}{2}\cos^2\theta_0 \right] $$

$$ F_{net} = \frac{P}{4c} (1-\cos\theta_0)^2 $$

Alternatively, taking negative arranging terms to match the form in the key :

$$ F_{net} = \frac{P}{2c} \left[ \frac{\sin^2\theta_0}{2} + \cos\theta_0 – 1 \right] \quad \text{} $$

Substituting $\sin^2\theta_0 = \frac{r^2}{r^2+d^2}$ and $\cos\theta_0 = \frac{d}{\sqrt{r^2+d^2}}$: