Problem 24.43

24.43 a)
We know that the magnitude of the electric and magnetic fields are related by the equation
$B_{0} = \frac{E_{0}}{c}$
So, $B_{0} = \frac{175 V/m}{3 \times 10^{8} m/s}$
$B_{0} = 5.83 \times 10^{-7} T$
The wave number k and wavelength are related by
$k = \frac{2\pi}{\lambda}$
$k = \frac{2\pi}{0.015 m}$
k = 419 rad/m
Also, the angular frequency and the wave number are related by
$\omega = kc$
$\omega = 1.26 \times 10^{11} rad/s$
We also know that since the Poynting vector is in the $\hat{x}$ direction and the electric field is in the $\hat{y}$ direction, then the magnetic field must be in the $\hat{z}$ direction since
$\vec{S} = \vec{E} \times \vec{B}$

b)
The average of the Poynting vector is given by
$S_{av} = \frac{E_{0} B_{0}}{2\mu_{0}}$
S_av = 40.6 W/m²

c)
If the sheet is perfectly reflecting then we have the equation for radiation pressure given by
$P_{r} = \frac{2S}{c}$
where the factor of two is due to the change in momentum from the incoming to the outgoing light. We know that the value of S is
S = 2 S_av
Putting it all together, we have
$P_{r} = 2.71 \times 10^{-7} N/m^{2}$

d)
The acceleration is given by
$a = \frac{F}{m}$
$a = \frac{P_{r} A}{m}$
$a = 4.07 \times 10^{-7} m/s^{2}$