Question 1

Multiple choiceMedium1 mark

A source emits light at a frequency of 5.000×1014 Hz5.000\times10^{14}\,\text{Hz} in its own rest frame. The source and observer move as shown in Fig. 1 at constant speeds, both much less than cc What frequency is measured by the observer?

sourceobserversuccessive light wavefrontslight propagationvs=3.00×10−3cvo=1.00×10−3c
Show answer and explanation

Answer: B. The source and observer move in the same direction, but the source is faster, so it approaches the observer at a relative speed of 0.00200c0.00200c The wavefronts are encountered more frequently, giving a fractional frequency increase of 0.002000.00200 Thus fobs=(5.000×1014)(1+0.00200)=5.010×1014 Hzf_{\text{obs}}=(5.000\times10^{14})(1+0.00200)=5.010\times10^{14}\,\text{Hz}

A: Uses the observer’s speed alone, predicting a 0.001000.00100 decrease in frequency.
C: Uses the source’s speed alone and ignores the observer’s motion.
D: Adds the speeds despite their being in the same direction, giving a fractional increase of 0.004000.00400

Remember: for motion in the same direction, the relative speed is the difference between the speeds.

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