Past exam of the physics 2 paper of the University of Oxford / 2016 resit / 8 / Solution

In vacuum, with possible sources $\rho$ and $\mathbf{J}$, Maxwell's equations areTheir integral forms, with compatible orientations and fixed surfaces, areCharge- and current-free space additionally sets $\rho =0$, $\mathbf{J}=0$. Vacuum constitutive constants alone do not imply that no charges exist.

Take the curl of the source-free Ampere-Maxwell law, use Faraday's law of induction and $\nabla \cdot \mathbf{B}=0$:ConsequentlyFor the specified complex plane wave, the wavevector is $k\hat{\mathbf{z}}$ and $\omega =ck$. Faraday's law of induction fixesThe physical fields are the real parts. They are in phase and their Poynting vector points along $+z$.

The instantaneous electric and magnetic energy densities are $u_E={ϵ}_0{E}^2/2$ and $u_B=B^2/(2{\mu }_0)$. Since $E=cB$, they are equal for this ideal travelling plane wave, and so are their cycle averages: $⟨u_E⟩/⟨u_B⟩=1$. For real peak amplitude $E_0$, each average is ${ϵ}_0{E}_0^2/4$. The Poynting vector and mean intensity satisfyIt represents transported energy per area per time, in $\mathrm{W}{\mathrm{m}}^{-2}$.

For the numerical moonlight estimate, adopt $L_{\odot }=3.828\times 10^{26}\mathrm{W}$ and lunar mean radius $R_M=1.7374\times 10^6\mathrm{m}$. These additional astronomical inputs give solar irradiance at the Moon of ${\mathcal{I}}_{\odot }=L_{\odot }/(4\pi d_{\odot }^2)\simeq 1361\mathrm{W}{\mathrm{m}}^{-2}$.^[ref][ref]

Under the simple isotropic point-reflector approximation, the intercepted power is $\pi R_M^2{\mathcal{I}}_{\odot }$, of which fraction $A=0.12$ is redistributed over $4\pi$ steradians. ThusThese are equivalent sinusoidal peak amplitudes; the corresponding root mean square fields are smaller by $\sqrt{2}$. Actual moonlight is broadband and incoherent.

A total reflected fraction alone does not specify the directional brightness of a full Moon. For a Lambertian sphere at full phase, with the same total reflectance $A$, integrating its radiance gives ${\mathcal{I}}_M=(2A/3){\mathcal{I}}_{\odot }(R_M/d_M{)}^2$, instead giving $E_0\simeq 1.30\mathrm{V}{\mathrm{m}}^{-1}$ and $B_0\simeq 4.32\mathrm{nT}$. If 0.12 is instead the geometric albedo, the full-phase formula is ${\mathcal{I}}_M=0.12{\mathcal{I}}_{\odot }(R_M/d_M{)}^2$, giving $E_0\simeq 1.59\mathrm{V}{\mathrm{m}}^{-1}$. NASA's table itself distinguishes Bond albedo from geometric albedo.^[ref] The numerical estimate therefore needs an explicit reflection convention; the first boxed values use the stated isotropic approximation.