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?
To find the peak value of the electric field associated with the light, we need to use the principle of mechanical equilibrium. When a beam of light is incident on the foil, it exerts radiation pressure on the surface of the foil, which causes it to bend. The foil reaches an equilibrium angle when the gravitational force acting on the foil is balanced by the radiation pressure.
1. Calculate the gravitational force acting on the foil:
The gravitational force can be calculated using the formula:
F_gravity = m * g
where m is the mass of the foil and g is the acceleration due to gravity (approximately 9.8 m/s^2). Substituting the given values:
F_gravity = (5.0 × 10^-5 kg) * (9.8 m/s^2) = 4.9 × 10^-4 N
2. Calculate the torque acting on the foil:
The torque can be calculated using the formula:
Torque = F_gravity * d * sin(theta)
where d is the distance from the point of suspension to the center of the foil, and theta is the equilibrium angle (2 degrees in this case). Since the foil is suspended along one edge, the distance d is equal to half the length of one side of the foil.
d = sqrt(A/4)
where A is the area of the square foil.
d = sqrt(5.0 × 10^-4 m^2 /4) = 0.010 m
Substituting the values:
Torque = (4.9 × 10^-4 N) * (0.010 m) * sin(2 degrees) = 5.1 × 10^-6 N*m
3. Calculate the radiation pressure exerted by the light:
The radiation pressure can be calculated using the formula:
P_radiation = Torque / A
where A is the area of the foil. Substituting the values:
P_radiation = (5.1 × 10^-6 N*m) / (5.0 × 10^-4 m^2) = 10.2 N/m^2
4. Find the peak value of the electric field associated with the light:
The radiation pressure is related to the electric field intensity by the formula:
P_radiation = (2 * E^2) / (c * u0)
where E is the peak value of the electric field, c is the speed of light (approximately 3 × 10^8 m/s), and u0 is the vacuum permeability (approximately 4π × 10^-7 T*m/A).
Rearranging the formula to solve for E:
E = sqrt((P_radiation * c * u0) / 2)
Substituting the values:
E = sqrt((10.2 N/m^2 * 3 × 10^8 m/s * 4π × 10^-7 T*m/A) / 2)
E = 4.03 × 10^2 N/C
Therefore, the peak value of the electric field associated with the light is 4.03 × 10^2 N/C.