Physics Circle home Student Login

Home›Chapters›Wave Motion PYQs›Doppler Effect›Question 60

JEE Main 2017 · Doppler Effect

Doppler Effect question from JEE Main 2017

Q60JEE Main 2017Medium

Doppler Effect

An observer is moving with half the speed of light towards a stationary microwave source emitting waves at frequency 10 GHz. What is the frequency of the microwave measured by the observer? (speed of light = 3 ×108 ms–1)

Show answer and solution

Answer: (d)

  1. For light use the relativistic Doppler formula: f′=f1+β1−βf' = f\sqrt{\dfrac{1 + \beta}{1 - \beta}} with β=0.5\beta = 0.5.
  2. f′=101.50.5=103≈17.3 GHzf' = 10\sqrt{\dfrac{1.5}{0.5}} = 10\sqrt{3} \approx 17.3\text{ GHz}
  3. So the answer is (d) 17.3 GHz.
Report a mistake

More questions on Doppler Effect