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Question
Three observers A, B and C measure the speed of light coming from a source to be νA, νBand νC. A moves towards the source and C moves away from the source at the same speed. B remains stationary. The surrounding space is vacuum everywhere.
(a) \[\nu_A > \nu_B > \nu_C\]
(b) \[\nu_A < \nu_B < \nu_C\]
(c) \[\nu_A = \nu_B = \nu_C\]
(d) \[\nu_B = \frac{1}{2}\left( \nu_A + \nu_C \right)\]
Solution
(c) \[\nu_A = \nu_B = \nu_C\]
(d) \[\nu_B = \frac{1}{2}\left( \nu_A + \nu_C \right)\]
Since the speed of light is a universal constant, \[\nu_A = \nu_B = \nu_C = 3 \times {10}^8\]
\[\nu_B = \frac{1}{2}\left( v_A + v_C \right)\]
This expression also implies that vA = vB = vC.
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