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 determine the peak value of the electric field associated with the light, we can use the concept of the torque on the suspended foil.
When light is incident perpendicular to the foil, it exerts a radiation pressure on the foil, causing it to bend. The torque due to this radiation pressure can be balanced by the torque due to the gravitational force acting on the foil.
The torque due to the radiation pressure is given by the equation:
τ = 2πIc, where τ is the torque, I is the intensity of light, and c is the speed of light.
The torque due to the gravitational force is given by the equation:
τ = mgl sin(θ), where τ is the torque, m is the mass of the foil, g is the acceleration due to gravity, and θ is the angle of deflection.
At equilibrium, these two torques are balanced:
2πIc = mgl sin(θ)
We can solve this equation to find the intensity of light, I:
I = (mgl sin(θ))/(2πc)
Now, we can calculate the peak value of the electric field using the equation:
I = (cε₀E₀²)/2, where I is the intensity, c is the speed of light, ε₀ is the vacuum permittivity, and E₀ is the peak value of the electric field.
Rearranging this equation gives us:
E₀² = (2I)/(cε₀)
Substituting the value of I we calculated earlier, we get:
E₀² = [(2(mgl sin(θ))/(2πc)] / (cε₀)
Finally, taking the square root of both sides gives us the peak value of the electric field, E₀:
E₀ = √[((2(mgl sin(θ))/(2πc)] / (cε₀)
Plugging in the given values:
m = 5.0 × 10^-5 kg
g = 9.8 m/s²
l (length of square) = √A = √(5.0 × 10^-4 m²) = 0.0224 m
θ = 2 degrees = 0.0349 radians
c = 3.00 × 10^8 m/s
ε₀ = 8.85 × 10^-12 C²/Nm²
We can now calculate the peak value of the electric field using the equation above.