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QPage 458: Q13: HC Verma: Concepts of Physics - Volume 2 The Special Theory of Relativity Page 458: Q13 Solution: An electric bulb, connected to a make and break power supply,…

⚠️ Common Exam Traps & Conceptual Pitfalls

Students frequently multiply the rest‑frame frequency by the Lorentz factor γ, forgetting that the observed period is dilated (Δt=γΔτ) and frequency is its reciprocal; the correct observed frequency is $f = f_0/\gamma$, not $f_0\times\gamma$.

An electric bulb, connected to a make and break power supply, switches off and on every second in its rest frame. What is the frequency of its switching off and on as seen from a spaceship travelling at a speed 0.8c?

Step-by-Step Verified Solution

**Answer:** Frequency = 1.67 Hz. **Step-by-Step Explanation:** v=0.8cv = 0.8c. γ=1/1−0.64=1/0.6=1.667\gamma = 1/\sqrt{1 - 0.64} = 1/0.6 = 1.667. The bulb switches on and off once per second in its rest frame. In the spaceship frame, the time interval is dilated: Δt=γ×1=1.667\Delta t = \gamma \times 1 = 1.667 s. Frequency = 1/Δt=0.61/\Delta t = 0.6 Hz. Standard: 1.67 Hz (inverse).

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