Doppler Effect for Light — Question 6

Question 6

Reflection from a receding mirror produces two Doppler factors

A plane mirror recedes along +x at $v=0.20c$. A laboratory wave of frequency $f=10.0\,\mathrm{GHz}$ propagates along +x, reflects backward, and returns. The mirror is massive enough that recoil is negligible; reflection preserves frequency in its rest frame.

1. Transform to the mirror frame and back to derive the returned frequency ratio.

2. Find the return frequency and compare the fractional shift with the small-speed approximation $-2\beta$.

Question 6 - Solution

Reflection from a receding mirror produces two Doppler factors

(a) Method and calculation

$$f_M=\gamma(1-\beta)f,\qquad f_R=\gamma(1-\beta)f_M.$$$$\boxed{f_R/f=\gamma^2(1-\beta)^2=\frac{1-\beta}{1+\beta}}.$$

(b) Interpretation and further calculation

$$\boxed{f_R=(2/3)(10.0)=6.66667\,\mathrm{GHz}}.$$$$\boxed{(f_R-f)/f=-1/3},\qquad-2\beta=-0.400.$$

Physical check

The two factors correspond to incident and reflected propagation. A one-way shift would underestimate the reflection effect.