Light as a Wave — Question 5

Question 5

The sign of a reflection coefficient determines a half-cycle shift

At normal incidence the electric-field reflection coefficient from a lossless dielectric is $r=(n_i-n_t)/(n_i+n_t)$. Compare reflection A from air ($n_i=1$) into glass ($n_t=1.50$) with reflection B from glass into air. Compare phases relative to each local incident field; incident intensities are equal.

1. Calculate both signed field coefficients and reflection phase changes.

2. Calculate both reflected intensity fractions and the relative reflection phase.

Question 5 - Solution

The sign of a reflection coefficient determines a half-cycle shift

(a) Method and calculation

$$r_A=(1-1.50)/(1+1.50)=\boxed{-0.20},\quad r_B=\boxed{+0.20}.$$$$\boxed{\phi_{r,A}=\pi,\quad\phi_{r,B}=0\pmod{2\pi}}.$$

(b) Interpretation and further calculation

$$R_A=|r_A|^2=R_B=|r_B|^2=\boxed{0.040}.$$$$\boxed{\phi_{r,A}-\phi_{r,B}=\pi\pmod{2\pi}}.$$

Physical check

An intensity measurement loses the sign of the reflected electric field; equal reflectances do not imply equal reflection phases.