A square piece of foil with area A = 5.0 × 10^-4 m^2 and mass m = 5.0 × 10^-5 kg is suspended along one edge and hangs under its own weight. When light is incident perpendicular to the foil,the foil bends and reaches an equilibrium angle of 2 degrees. If the light is completely reflected by the foil, what is the peak value of the electric field associated with the light?
The answer is 62 kV/m.
I got...
Torque (by radiation) = F dot R = A*P dot R = A*P cos(theta) R
P = 0.5 * (electric constant) * E^2
Torque (by gravity) = -mg sin(theta)
Torque (by radiation) + Torque (by gravity) = 0
Then i solved for E. Which gives me a wrong answer. I don't know why.
To find the peak value of the electric field associated with the reflected light, we can use the concept of the bending of the foil.
When light is incident perpendicular to the foil, it exerts a pressure on the foil due to the transfer of momentum. This pressure causes the foil to bend until it reaches an equilibrium angle.
We can start by calculating the force exerted by the light on the foil. The force is given by the expression:
F = P/A
where F is the force, P is the pressure, and A is the area of the foil.
The pressure exerted by the light is equal to the momentum change per unit time due to reflection, divided by the area of the foil.
Since the light is completely reflected, the momentum change per unit time is twice the momentum of the incident light.
The momentum of light is given by the expression:
p = E/c
where p is the momentum, E is the energy of the light, and c is the speed of light.
The energy of the light is related to the electric field strength by the expression:
E = c * ε_0 * E^2 / 2
where E is the electric field strength and ε_0 is the permittivity of free space.
Substituting the expressions for momentum and energy into the equation for the force, we get:
F = (2 * E/c) / A
Now, we consider the torque on the foil due to the weight of the foil. The torque is given by the expression:
τ = m * g * L
where τ is the torque, m is the mass of the foil, g is the acceleration due to gravity, and L is the length of the suspended edge.
At equilibrium, the torque due to the weight is balanced by the torque due to the force exerted by light. Therefore, we have:
F * L = τ
Substituting the expressions for force and torque, we get:
[(2 * E/c) / A] * L = m * g * L
Canceling out L on both sides, we get:
(2 * E/c) / A = m * g
Simplifying, we have:
E = (m * g * c * A) / (2 * ε_0)
Finally, substituting the given values:
m = 5.0 × 10^-5 kg
g = 9.8 m/s^2
c = 3.0 × 10^8 m/s
A = 5.0 × 10^-4 m^2
ε_0 = 8.85 × 10^-12 C^2/(N·m^2)
We can now plug these values into the formula to find the peak value of the electric field associated with the reflected light.