Physics Solution – Light Pressure on Plates

Solution: Force of Light Pressure

Figure 1: Diagram showing numbered momentum flux pathways (Corrected: Ray 2 Left, Ray 6 Right).

1. Momentum Flux Analysis

We calculate the force exerted on the plates by analyzing the change in momentum of the light beams. Force is defined as the rate of change of momentum: $$\vec{F} = \vec{\Phi}_{\text{in}} – \vec{\Phi}_{\text{out}}$$ where $\vec{\Phi}$ represents the momentum flux (Power/c) vector.

Input Flux:

Ray 1 (Incident): $\vec{\Phi}_{in} = I(-\hat{j})$

Output Fluxes (see numbered rays):

Ray 2 (Reflected from Top): Intensity $\rho I$, Direction $-\hat{i}$ (Left).
Ray 4 (Transmitted through Bottom): Intensity $(1-\rho)^2 I$, Direction $-\hat{j}$ (Down).
Ray 6 (Secondary Reflected from Top): Intensity $\rho[\rho(1-\rho)I]$, Direction $+\hat{i}$ (Right).
Ray 7 (Secondary Transmitted from Top): Intensity $(1-\rho)[\rho(1-\rho)I]$, Direction $+\hat{j}$ (Up).

2. Calculation of Force Components

X-Component ($F_x$):

The net force in the x-direction depends on the difference between input and output x-momentum.

$$ \Phi_{\text{in}, x} = 0 $$ $$ \Phi_{\text{out}, x} = (\text{Ray 2}) + (\text{Ray 6}) $$ $$ \Phi_{\text{out}, x} = (-\rho I) + (\rho^2(1-\rho)I) $$ $$ \Phi_{\text{out}, x} = -I\rho (1 – \rho(1-\rho)) = -I\rho(1 – \rho + \rho^2) $$

Thus, the force $F_x$ is:

$$ F_x = \Phi_{\text{in}, x} – \Phi_{\text{out}, x} $$ $$ F_x = 0 – [-I\rho(1 – \rho + \rho^2)] $$ $$ F_x = +I\rho(1 – \rho + \rho^2) $$

The positive sign indicates the net x-force is towards the positive x-axis (Right).

Y-Component ($F_y$):

Similarly, for the y-direction:

$$ \Phi_{\text{in}, y} = -I \quad (\text{Ray 1}) $$ $$ \Phi_{\text{out}, y} = (\text{Ray 4}) + (\text{Ray 7}) $$ $$ \Phi_{\text{out}, y} = -(1-\rho)^2 I + \rho(1-\rho)^2 I $$ $$ \Phi_{\text{out}, y} = I(1-\rho)^2 (\rho – 1) = -I(1-\rho)^3 $$

Thus, the force $F_y$ is:

$$ F_y = \Phi_{\text{in}, y} – \Phi_{\text{out}, y} $$ $$ F_y = -I – [-I(1-\rho)^3] = -I [ 1 – (1-\rho)^3 ] $$

The negative sign indicates the force component is towards the negative y-axis.

3. Resultant Direction

We have a positive x-component ($F_x$) and a negative y-component ($F_y$). This places the resultant force vector in the 4th Quadrant. The question asks for the direction from the x-axis towards the negative y-axis. This corresponds exactly to the angle $\theta$ in the 4th quadrant measured clockwise from the positive x-axis.

$$ \tan \theta = \frac{|F_y|}{|F_x|} $$ $$ \tan \theta = \frac{I [ 1 – (1-\rho)^3 ]}{I \rho (1 – \rho + \rho^2)} $$ $$ \tan \theta = \frac{1 – (1-\rho)^3}{\rho(1 – \rho + \rho^2)} $$

Therefore, the direction is:

$$ \theta = \tan^{-1} \left\{ \frac{1 – (1-\rho)^3}{\rho(1 – \rho + \rho^2)} \right\} $$